The total cost C (in dollars) to participate in a ski club is given by the literal equation C=85x+60, where x is the number of ski trips you take.

a. Solve the equation for x.

x=

b. How many ski trips do you take if you spend a total of $315? $485?

ski trips costs $315.

ski trips costs $485.

Answer:

  1, 5, 8

Step-by-step explanation:

1 is a vertical angle with angle 4, so is congruent.

5 is an alternate interior angle with angle 4, so is congruent.

8 is a corresponding angle with angle 4, so is congruent.

To determine the angles that are congruent to angle 4, we need to find angles with the same measure.

In geometry, two angles are congruent if they have the same measure. To determine which angles are congruent to angle 4, we need to find other angles that have the same measure as angle 4.

Let's assume angle 4 measures x degrees. In this case, any angle that also measures x degrees would be congruent to angle 4.

if angle 4 measures 60 degrees, any other angle that measures 60 degrees.

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To estimate the result of 542,817 - 27,398, we can round to the nearest ten thousand and perform the subtraction, yielding an estimation of 510,000.

The student wants to estimate the result of 542,817 - 27,398. One way of doing this is by rounding to the nearest ten thousand. In this case, 542,817 becomes 540,000 and 27,398 becomes 30,000.

Now, perform the calculation: 540,000 - 30,000 = 510,000. Hence, the estimated result of 542,817 - 27,398 is roughly 510,000.

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Answer:

She have both activities again on 20th day from now.

Step-by-step explanation:

Given :Ronna has volleyball practice every 4 days

         Ronna also has violin lessons every 10 days.

To Find : When will she have both activities again on the same day?

Solution:

We will find the LCM of 4 and 10

2  |   4 ,10

2  |   2 , 5

5  |   1  , 5

   |   1  ,  1

So, [tex]LCM = 2\times 2 \times 5[/tex]

[tex]LCM =20[/tex]

Hence she have both activities again on 20th day from now.

The measure of angle ABC in the figure is 53 degrees.

Option D) 53° is the correct answer.

From the figure in the image;

Angle ABC and angle CBD are complementary angles, hence, their sum equals 180 degrees.

From the figure:

Angle ABC = ( 5x + 3 )

Angle CBD = ( 3x + 7 )

First, we solve for x:

Since the two angles are complementary:

Angle ABC + Angle CBD = 90

( 5x + 3 ) + ( 3x + 7 ) = 90

5x + 3x + 3 + 7 = 90

8x + 10 = 90

8x = 90 - 10

8x = 80

x = 80/8

x = 10

Now, we find the measure of angle ABC:

Angle ABC = ( 5x + 3 )

Plug in x = 10:

Angle ABC = 5(10) + 3

Angle ABC = 50 + 3

Angle ABC = 53°

Therefore, angle ABC measures 53 degrees.

The correct option is D) 53°

Answer:

5%.

Step-by-step explanation:

We have been given that a pharmacist attempts to weigh 120 mg of codeine sulfate on a balance with a sensitivity requirement of 6 mg.

To find the maximum potential error, we need to figure out 6 mg is what percentage of 120 mg.

[tex]\text{The maximum potential error}=\frac{6}{120}\times 100\%[/tex]

[tex]\text{The maximum potential error}=0.05\times 100\%[/tex]

[tex]\text{The maximum potential error}=5\%[/tex]

Therefore, the maximum potential error is 5%.

Divide:

62.98 / 185.95 = 0.338693

Multiply the decimal by 100:

0.338693 x 100 = 33.8693%

Round the answer as needed.

Answer: [tex]d) 2.0*10^2[/tex]kilo gram

Step-by-step explanation:  

To know the greatest mass we have to convert all these measurements to the same unit, the gram. For that we have to state the next units:

[tex]mili = 10^-^3\\micro = 10^-^6\\pico = 10^-^1^2\\kilo= 10^3[/tex]

Now we can convert the measurements to the same unit:

[tex]b)5.0*10^1^2*10^-^1^2gram=5.0gram\\d)2.0*10^2*10^3gram=2.0*10^5gram\\c)9.0*10^9*10^-^6gram=9.0*10^3gram\\a)2.0*10^6*10^-^3gram=2.0*10^3gram[/tex]

Comparating the values, we can see that the option d) is the greatest mass.

Answer:

The greatest mass among the following measurements is:

[tex]2*10^{2}kg = 200kg[/tex]

Step-by-step explanation:

I am going to convert each one of these measurements to kilogram, then the biggest value in kg is the greatest mass.

a) [tex]2*10^{6}mg[/tex]

Each kg has [tex]10^{6} kg[/tex].

So

1 kg - [tex]10^{6} kg[/tex]

x kg - [tex]2*10^{6}mg[/tex]

x = 2.

So

[tex]2*10^{6}mg = 2kg[/tex]

b)[tex]5*10^{12}pg[/tex]

Each kg has [tex]10^{15}pg[/tex].

So

1 kg - [tex]10^{15}pg[/tex]

x kg - [tex]5*10^{12}pg[/tex]

[tex]x = 5*10^{-3}kg[/tex]

This is a smaller mass than a)

c) [tex]9.0*10^{9} mcg[/tex]

Each kg has  [tex]10^{9}mcg[/tex].

So

1 kg - [tex]10^{9}mcg[/tex]

x kg - [tex]9*10^{9} mcg[/tex]

x = 9kg.

[tex]9.0*10^{9} mcg = 9kg[/tex]

For now, this is the greatest mass.

d) [tex]2*10^{2}kg = 200kg[/tex]

This is the greatest mass.

Answer:

solved

Step-by-step explanation:

Calculating i found the value of [tex]\frac{1}{128}[/tex]= 0.0078125

a.therefore the approximate valve of  [tex]\frac{1}{128}[/tex] (correct to three       decimal places) = 0.008

b. there are three significant digits in my approximation.

c. the absolute error = | exact value- approximate value |

   =|0.0078125- 0.008|= |-0.0001875| = 0.0001875

d.relative error= [tex]\frac{ absolute\ error}{exact\ value}[/tex]

    = [tex]\frac{0.0001875}{0.0078125}[/tex]= 0.024

Answer:

The vector equation of the line is

l(t)= "-3,3,-5" + t "-2,-4,-4", where t is a parameter.

Answer:

Intersection of collection of any closed set is a closed set.    

Step-by-step explanation:

Let F be a collection of arbitrary closed sets and let [tex]B_i[/tex] be closed set belonging to F.

We define a closed set as the set that contains its limit point or in other words it can be described that the complement or not of a closed set is an open set.

Thus, we can write R as

[tex]R =\bigcap\limits_{B_i \in F }^{} B_i[/tex]

Now, applying De-Morgan's Theorem, we have

[tex]R^c = (\bigcap\limits_{B_i \in F }^{} B_i)^c[/tex]

[tex]R^c = \bigcup\limits_{B_i \in F }^{} B_i^c[/tex]

Since we knew[tex]B_i[/tex] are closed set, thus, [tex]B_i^c[/tex] is an open set.

We also know that union of all open set is an open set.

Thus, [tex]R^c[/tex] is an open set.

Thus, R is a closed set.

Hence, the theorem.  

The intersection of any collection of closed sets is indeed closed. This can be proven using the properties of complements in topology and by employing a contradiction approach where the assumption that the intersection is not closed leads to a contradiction, hence proving it must be closed.

The question addresses whether the intersection of any collection of closed sets is closed. In topology, a closed set is one where its complement (the elements not in the set) is open. One way to approach the proof is considering De Morgan's Laws, which in topology state that the intersection of closed sets is the complement of the union of their complements, which are open sets. Since the union of open sets is open, it follows that the complement (the intersection of our original sets) is closed.

For example, consider two closed intervals on the real line, A and B. The intersection, A B, would be the set of all points that are in both A and B. The fact that both A and B contain their boundary points ensures that their intersection also contains boundary points, maintaining closedness.

Proof strategy involves contradiction: assume the intersection is not closed. This would imply that its complement is not open, violating the definition that complements of closed sets are open, thus leading to a contradiction and proving the proposition is true.

Answer:

Angle between the two vectors is 20.7718°.

Step-by-step explanation:

The given two vectors are x = (1,2,3,4) and y = (4,2,4,5)

Now the dot product between two vector is defined as

x.y = ║x║║y║cosθ

Now, x.y( dot product of x and y) = 1×4 + 2×2 + 3×4 + 4×5 = 40

║x║= [tex]\sqrt{1+4+9+16}[/tex] = [tex]\sqrt{30}[/tex]

║y║= [tex]\SQRT{16+4+16+25} = \sqrt{61}[/tex] '

Thus, we get

40 = [tex]\sqrt{30}[/tex]×[tex]\sqrt{61}[/tex]×Cos θ

Thus, Cos θ = [tex]\frac{40}{\sqrt{30}\sqrt{61}}[/tex]

Cos θ = 0.935

θ = [tex]Cos^{-1} (0.935)[/tex]

θ = 20.7718°

Answer:

The dimensions are 68 meters by 42 meters.

Step-by-step explanation:

Let the length be x

Let the width be y

Perimeter = 220 meter

So, [tex]2x+2y=220[/tex]

or [tex]x+y=110[/tex] or [tex]x=110-y[/tex]    ...........(1)

Area = 2856 square meter

So, [tex]xy=2856[/tex]     ............(2)

Substituting the value of x from (1) in (2)

[tex](110-y)y=2856[/tex]

=> [tex]110y-y^{2}=2856[/tex]

=> [tex]y^{2}-110y+2856=0[/tex]

Solving this equation, we get roots as y = 68 and y = 42

So, if we put y = 68, we get x = 42

If we put y = 42, we get x = 68

As length is longer than width, we will take length = 68 meters and width = 42 meters.

Hence, the dimensions are 68 meters by 42 meters.

The probability that a randomly selected bacterium that's currently alive will still be alive after one week is 1/3. The calculation is done by dividing the number of bacteria that can stay alive for at least 30 days by the total number of bacteria that are currently alive.

The question pertains to the concept of probability in mathematics and it's asking for the likelihood that a randomly chosen bacteria from a test tube would still be alive after one week. Given that 15 bacteria are alive (5 will stay alive for at least 30 days, and 10 will die on the second day), and you have chosen a bacterium that is currently alive. After one week (7 days), only the 5 that can stay alive for at least 30 days will still be alive.

As a result, the probability that the chosen bacterium will still be alive after one week can be calculated by dividing the number of bacteria that can stay alive for at least 30 days (5) by the total number of currently alive bacteria (15).

Probablitiy = number of favourable outcomes / Total number of outcomes
Therefore, the probability that it will still be alive after one week is 5 / 15 = 1/3.

Therefore, the answer is (a) 1/3.

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Answer:

True

Step-by-step explanation:

Let B={[tex]v_1, v_2,\cdots,v_k,0[/tex]} a set of vectors containing the zero vector.

B is linear dependent if exist scalars [tex]a_0,a_1, a_2,\cdots,a_k[/tex] not all zero such that [tex]a_0*0+a_1*v_1+\cdots+a_k*v_k=0[/tex].

Observe that if [tex]a_1=a_2=\cdots=a_k=0[/tex] and [tex]a_0=\lambda\neq 0[/tex] then

[tex]\lambda*0+0*v_1+\cdots+0*v_k=0[/tex].

Then B is linear dependent.

Final answer:

True, any set of vectors that contains the zero vector is linearly dependent because a nontrivial combination of the vectors (including the zero vector scaled by any scalar) can produce the null vector.

Explanation:

True. Any set of vectors containing the zero vector is linearly dependent. This is because the null vector, which has all components equal to zero (0î + 0ç + 0k) and therefore no length or direction, can always be represented as a linear combination of other vectors in the set by simply scaling it by any scalar. In more technical terms, for a set of vectors that includes the zero vector, you can find a nontrivial combination (not all scalars being zero) such that when these scalars are applied to their respective vectors, their sum equals the zero vector. This is the very definition of linear dependency. For instance, if you consider any scalar 'a' not equal to zero, a∗0 + 0∗0 + ... + 0∗0 = 0, which demonstrates linear dependence because not all the scalars are zero (namely 'a' is not).

Linear independence would require that the only way to represent the zero vector as a linear combination of the set of vectors is to have all scalars that multiply each vector in the combination be zero. Since the set includes the zero vector itself, this condition is violated, and the set is dependent.

Answer:

a) [tex]636_{7}[/tex] and  [tex]641_{7}[/tex]

b) [tex]11111_{2}[/tex] and  [tex]100001_{2}[/tex]

c) [tex]554_{6}[/tex] and  [tex]1000_{6}[/tex]

d)[tex]44_{5}[/tex] and  [tex]101_{5}[/tex]

e)[tex]3333_{4}[/tex] and  [tex]10001_{4}[/tex]

f) [tex]404_{6}[/tex] and  [tex]410_{6}[/tex]

Step-by-step explanation:

The logics followed in order to find them:

Base numbers work exactly the same as base 10 numbers (the ones we use on a daily basis). Take for example, in a decimal system, we have 10 digits available:

0,1,2,3,4,5,6,7,8,9.

when counting, when we get to the 9, we start over again, but writting a 1 to the left.

0,1,2,3,4,5,6,7,8,9,10,11,12....20,21,22,23....,90,91,92,93,94,95,96,97,98,99,100

when reaching 99, we have no more digits to use, so we start the count again and go from 99 to 100.

The same works with any other base number, take for example the base 7 numbet. When counting in base 7, we only have 7 digits available: (0,1,2,3,4,5,6) So when we re0_{7},ach the digit 6, we go immediately to 10, like this:

[tex]0_{7},1_{7},2_{7},3_{7},4_{7},5_{7},6_{7},10_{7},11_{7},12_{7},13_{7},14_{7},15_{7},16_{7},20_{7}...[/tex]

notice how it went from 6 to 10 and from 16 to 20. This is because in base 7, there is no such thing as digits from 7 to 9, so we don't have any other option to go directly to 20.

So, on part A) if we were working with decimal numbers, the  previous value for 640 would be 639, but notice that in base 7, there is no such thing as the digit 9, so the greatest digit we can use there would be 6, therefore, the previous value for the [tex]640_{7}[/tex] number would be [tex]636_{7}[/tex]. The next number would be [tex]641_{7}[/tex] because the 1 does exist for a base 7 system.

The same logics is followed for the rest of the problems.

One less than the given number is known as preceding number whereas one more than the given number is the succeeding number.

The logics followed in order to find them:

Base numbers work exactly the same as base 10 numbers (the ones we use on a daily basis). Take for example, in a decimal system, we have 10 digits available:

0,1,2,3,4,5,6,7,8,9.

When counting, when we get to the 9, we start over again, but writing a 1 to the left.

0,1,2,3,4,5,6,7,8,9,10,11,12....20,21,22,23....,90,91,92,93,94,95,96,97,98,99,10

When reaching 99, we have no more digits to use, so we start the count again and go from 99 to 100.

The same works with any other base number, take for example the base 7 number. When counting in base 7, we only have 7 digits available: (0,1,2,3,4,5,6) So when 0_{7},each the digit 6, we go immediately to 10, like this:

It went from 6 to 10 and from 16 to 20. This is because in base 7, there is no digits from 7 to 9, so we have directly to 20.

So, if working with decimal numbers, the previous value for 640 would be 639, but notice that in base 7,

There is no such thing as the digit 9, so the greatest digit we can use there 6,

Therefore, the previous value[tex]636_7[/tex] for the number would be [tex]636_7[/tex] . The next number would be [tex]632_7[/tex] because the 1 does exist for a base 7 system.

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Answer:

Multiplication is commutative.          

Step-by-step explanation:

Let n,m be two natural numbers.

We have to show that [tex]n\times m=m\times n[/tex]

We will prove this by induction

For n = 1, we have [tex]n\times 1 = 1\times n = n[/tex], which is true.

Let the given statement be true for n-1 natural numbers, that is,

[tex](n-1)\times m = m\times (n-1)[/tex]

Now, we have to show that it is true for n natural numbers.

[tex]nm = (n-1)m + m = m(n-1) + m[/tex]

which is equal to

[tex](m-1)(n-1) + (n-1) + m[/tex]

[tex]= (m-1)(n-1) + (m-1) + n[/tex]

[tex]= n(m-1) + n[/tex]

[tex]= (m-1)n + n[/tex]

[tex]= mn[/tex]

Hence, by principal of mathematical induction, the given statement is true for all natural numbers.

Answer:

17. C1 = 1    and    C2 = 0

18. C1 = 4/3    and    C2 = -1/3

Step-by-step explanation:

See it in the picture

Answer:

[tex]v_{1}.v_{2} = 0.804[/tex]

In terms of the class, the dot product represents the weighed class average.

Step-by-step explanation:

The two vectors are:

-[tex]v_{1}:[/tex] The weight of each of the semester's exams.

[tex]v_{1} = (10%, 15%, 25%, 50%)[/tex]

In decimal:

[tex]v_{1} = (0.10, 0.15, 0.25, 0.50)[/tex]

-[tex]v_{2}:[/tex] The class average on each of the exams

In decimal:

[tex]v_{2} = (0.75, 0.91, 0.63, 0.87)[/tex]

-----------------------

Dot product:

Suppose there are two vectors, u and v

u = (a,b,c)

v = (d,e,f)

There dot product between the vectors u and v is:

u.v = (a,b,c).(d,e,f) = ad + be + cf

------------------

So

[tex]v_{1}.v_{2} = (0.10, 0.15, 0.25, 0.50).(0.75, 0.91, 0.63, 0.87) = 0.10*0.75 + 0.15*0.91 + 0.25*0.63 + 0.50*0.87 = 0.804[/tex]

[tex]v_{1}.v_{2} = 0.804[/tex]

In terms of the class, the dot product represents the weighed class average.

Answer:

[tex]C^{102}_{100}=5151[/tex]

Step-by-step explanation:

Let's imagine the following situation, if we want to distribute 100 coins between three pirates we could represent this situation with a line arrangement. For example if we had7 coins and 3 pirates one possible distribution of coins would be given by  CC|CCCC|C, the C's represent coins and the bars the boundaries between two pirates, for the particular line arrangement shown, we have that pirate A has 2 coins, B has 4 coins and C has a single coin. Another possible arrangement is,

|CCC|CCCC, where pirate A has no coin, pirate B has 3 coins and C has 7 coins. If we take notice of the fact that the arrangement representing a distribution is composed of 9 elements, that is 7 C's and 2 | (bars), then a way to make an arrangement would be to fill 9 empty boxes with our available coins and bars in all the possible ways. This means that if we first choose to fill 7 out of 9 boxes with  coins then the number of possible combinations is [tex]C^{9}_7=\frac{9!}{7!(9-7)!}36[/tex]. In general if we want to distribute n elements in k boxes, where the boxes can either be filled with any number of elements (including 0 number of elements), we have that the number of possible distributions will be [tex]C^{n+k-1}_{n}=\frac{(n+k-1)!}{n!(k-1)!}[/[tex], where we used the fact that we need k-q bars to represent k boxes. Thus pirate A can choose from [tex]C^{102}_{100}=5151[/tex] possible divisions.

Bonus:

If every pirate wants to have the maximum number of coins possible without being executed, here's how pirate A has to divide the coins in order to keep the largest amount of coins.

We have to think backwards to figure this out. Imagine pirate A was executed and there are only two remaining players. Pirate B should propose to keep all the coins, pirate C could oppose but pirate B's vote would break the vote and keep all the loot. Pirate A, B and C are all aware of this, so pirate A should propose to keep 99 coins and give the remaining gold piece to pirate C, Pirate B will of course oppose the division, but pirate C should accept because if not he would get no coins. Thus the division would be.

A: 99 coins

B: 0 coins

C: 1 coins

Answer:

(C) Of the lost items, 11% cost between $100 and $200

Step-by-step explanation:

On the Y-axis is represented the frequency of lost items found.

In this case in the form of a percentage.

On the X-axis is represented the range of prices

See picture attached.

So, this means that 11% of the found items have an estimated price between $100 and $200

Answer: There are 104 silvery minnow population in the Rio Grande River.

Step-by-step explanation:

Since we have given that

Number of tagged silvery minnows = 54

Number of captured silvery minnows = 62

It includes 12 tagged silvery minnows.

So, Number of silvery minnows without tags = 62 -12 =50

So, Good estimate of the silvery minnow population in the Rio Grande River would be

[tex]54+50\\\\=104[/tex]

Hence, there are 104 silvery minnow population in the Rio Grande River.

Answer:

Step-by-step explanation:

[tex]$$a. Sam had pizza last night and Chris finished her homework.\\p\wedge q\\\\$b. Chris did not finish her homework and Pat watched the news this morning.$\\\neg q \wedge r\\\\$c. Sam did not have pizza last night or Chris did not finish her homework.$\\\neg p \vee \neg q\\\\[/tex]

[tex]$$d. Either Chris finished her homework or Pat watched the news this morning, but not both.$\\q\vee r\\\\$e. If Sam had pizza last night then Chris finished her homework.$\\p \rightarrow q\\\\$f. Pat watched the news this morning only if Sam had pizza last night.$\\p\leftrightarrow r\\\\[/tex]

[tex]$$g. Chris finished her homework if Sam did not have pizza last night.$\\\neg p \rightarrow q\\\\$h. It is not the case that if Sam had pizza last night, then Pat watched the news this morning.$\\\neg (p\rightarrow r)\\\\[/tex]

[tex]$$i. Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning.$\\(\neg p \wedge q) \Rightarrow r\\\\ $j. q\Rightarrow r$\\Chris finished her homework implies that Pat watched the news this morning.\\\\[/tex]

[tex]$$k. p \Rightarrow (q \wedge r)$\\Sam has pizza last night implies that Chris finished her homework and Pat watched the news this morning.$\\\\[/tex]

[tex]$$l. \neg p \Rightarrow (q \vee r)$\\Sam did not have pizza last night implies that Chris finished her homework or Pat watched the news this morning.$\\\\[/tex]

[tex]$$m. r \Rightarrow (p \vee q)$\\Pat watched the news this morning implies that Sam had pizza last night or Chris finished her homework$[/tex]

Final answer:

Logical formulas for the given statements about Sam, Chris, and Pat are provided using logical operators such as 'and', 'or', 'not', and 'implies'. The logical symbols ∧, ∨, ¬, and ⇒ are utilized to form the statements. Some formulas are also expressed in words to match their logical predictions with linguistic intuitions.

Explanation:

The logical formulas for the statements given about Sam, Chris, and Pat using p, q, and r are as follows:

a) Sam had pizza last night and Chris finished her homework: p ∧ q

b) Chris did not finish her homework and Pat watched the news this morning: ¬q ∧ r

c) Sam did not have pizza last night or Chris did not finish her homework: ¬p ∨ ¬q

d) Either Chris finished her homework or Pat watched the news this morning, but not both: q ⊕ r

e) If Sam had pizza last night then Chris finished her homework: p ⇒ q

f) Pat watched the news this morning only if Sam had pizza last night: r ⇒ p

g) Chris finished her homework if Sam did not have pizza last night: ¬p ⇒ q

h) It is not the case that if Sam had pizza last night, then Pat watched the news this morning: ¬(p ⇒ r)

i) Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning: (¬p ∧ q) ⇒ r

Now let's express some formulas in words:

j) q ⇒ r: If Chris finished her homework, then Pat watched the news this morning.

k) p ⇒ (q ∧ r): If Sam had pizza last night, then Chris finished her homework and Pat watched the news this morning.

l) ¬p ⇒ (q ∨ r): If Sam did not have pizza last night, then Chris finished her homework or Pat watched the news this morning.

m) r ⇒ (p ∨ q): If Pat watched the news this morning, then Sam had pizza last night or Chris finished her homework.

Answer with Step-by-step explanation:

We are given that sum of three vectors in [tex]R^3[/tex] is zero.

We have to tell and explain that if sum of three vectors in [tex]R^3[/tex] is zero then they must lie in the same plane or not.

We know that if three or more vectors lie in the same then they are coplanar.

If three vectors are co-planar then their scalar product is zero.

According to given condition

Let (x,0,0), (-x,0,0) and (0,0,0)

[tex]\vec{u}=x\hat{i}[/tex]

[tex]\vec{v}=-x\hat{i}[/tex]

[tex]\vec{w}=0[/tex]

Sum of three vectors=[tex](x-x+0)\hat{i}+(0+0+)\hat{j}+(0+0+0)\hat{k}=0[/tex]

[tex]u\cdot (v\times w)=\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}[/tex]

[tex]u\cdot (v\times w)=\begin{vmatrix}x&0&0\\-x&0&0\\0&0&0\end{vmatrix}[/tex]

When all elements of one row or column are zero of square matrix A then det(A)=0

[tex]u\cdot (v\times w)=0[/tex]

Therefore, vectors u,v and w are co-planar.

Hence, if the sum of three vectors in [tex]R^3[/tex] is zero then they must lie in the same plane.

If the sum of three vectors in three-dimensional space is zero, they typically lie in the same plane, forming a triangle with their magnitudes and directions. They must also be linearly dependent since their linear combination can produce the zero vector.

If the sum of three vectors is zero in R^3 (three-dimensional space), it is a common situation for these vectors to lie in the same plane. This condition implies that they can be represented as three sides of a triangle taken in sequence in terms of magnitude and direction. If we consider two vectors in sequence, such as AB and BC, the third vector must be the closing side CA, taken in the opposite direction, in order to satisfy the triangle law of vector addition. In this case, the sum of vectors AB + BC + CA equals zero.

Furthermore, vectors are homogeneous when added, which means that they must have the same nature and dimension. However, three vectors can still form a linearly dependent set in three-dimensional space even if they lie in the same plane. For instance, if we take vectors a, b, and c, and if a linear combination of these vectors yields the zero vector (i.e., if we can find scalar coefficients x, y, and z such that xa + yb + zc = 0), then these vectors are not linearly independent and must lie in the same plane.

Answer : (a) 5040  b) 720  c) 288

Step-to step explanation:-

Given : There are 3 Computer Science, 2 Mathematics, and 2 Sociology textbooks.

Total books = 7

a) The number of ways the books can be arranged in any order is given by :-

[tex]7!=7\times6\times5\times4\times3\times2=5040[/tex]

b) We consider all computer books as one thing, then, the total number of things to arrange = 1+2+2=5

Now, the number of ways to arrange books such that Computer Science books must be together, but all other books can be arranged in any order:-

[tex]3!\times5!\\\\=6\times120=720[/tex]

c) We consider all computer books as one thing and math book as one thing, then, the total number of things to arrange = 1+1+2=4

Now, the number of ways to arrange books such that Computer Science books must be together, but all other books can be arranged in any order:-

[tex]4!\times 3!\times2!=24\times6\times2=288[/tex]

Answer:

B. It is greater than the median.

Step-by-step explanation:

If n is an odd, prime integer and 10<n<19, then the true statements about the mean of all possible values of n are:

B. It is greater than the median.

Here, n can be 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18.

But as n is odd , it can be 11, 13, 15 and 17.

And also it is given that n is prime, so we will cancel out 15.

Now we are left with 11, 13 and 17.

Their mean is = [tex]\frac{11+13+17}{3}[/tex] = 13.66

Median is 13, that is smaller than 13.66.

So, we can see that the mean is greater than median.

Answer:

(a) r = 57.628

    θ = 51.34°

(b) r = 0.711

    θ = 38.65°

Step-by-step explanation:

(a) Let's assume

 z = (5 + 4)(4 +5j)

    = 9 x (4+5j)

    = 36 + 45j

magnitude of z can be given by

[tex]\left | z \right |\ =\ r\ =\sqrt{36^2+45^2}[/tex]

               => r =57.628

angle of z can be given by,

     [tex]tan\theta\ =\ \dfrac{45}{36}[/tex]

[tex]=>\ tan\theta=\ \dfrac{5}{4}[/tex]

[tex]=>\ \theta\ =\ tan^{-1}\dfrac{5}{4}[/tex]

              θ = 51.34°        

(b) Let's assume

[tex]z =\ \dfrac{(5 +4j)}{5+4}[/tex]

    [tex]=\ \dfrac{(5+4j)}{9}[/tex]

    [tex]=\ \dfrac{5}{9}+\dfrac{4i}{9}[/tex]

magnitude of z can be given by

[tex]\left | z \right |=\ r=\sqrt{(\dfrac{5}{9})^2+(\dfrac{4}{9})^2}[/tex]

             [tex]=> r =\dfrac{\sqrt{41}}{9}[/tex]

              => r = 0.711

angle of z can be given by,

[tex]\ tan\theta\ =\ \dfrac{\dfrac{4}{9}}{\dfrac{5}{9}}[/tex]

[tex]=>\ tan\theta=\ \dfrac{4}{5}[/tex]

[tex]=>\ \theta\ =\ tan^{-1}\dfrac{4}{5}[/tex]

                => θ = 38.65°        

Answer: [tex]\sqrt{14}[/tex]

Step-by-step explanation:

The given notation is not clear enough, but I will asume that the question is: "Wich is the vector lenght of the vector [tex](2,3,1)[/tex] in a 3 dimensional euclidean space"

Vector lenght of a vector [tex](a,b,c)[/tex] is given by the next formula:

║[tex](a,b,c)[/tex]║=[tex]\sqrt{a^2+b^2+c^2}[/tex]

In this case we have a=2, b=3 and c=1. We should substitute these values in the formula given before to obtain:

║[tex](2,3,1)[/tex]║=[tex]\sqrt{2^2+3^2+1^2}=\sqrt{14}=3.74165[/tex]

Wich is the answer to the problem.  

Answer: Probability of trucks having both defects is 4%.

Step-by-step explanation:

Since we have given that

Probability of trucks having defective brake linings only = 6%

Probability of trucks having defective steering only = 3%

Probability of trucks having neither defect = 87%

Probability of trucks having either defect - 100 - 87 = 13%

Let the probability of trucks having both defects be 'x'.

P(defective brake lining) = P(B) = 0.06+x

P(defective steering) = P(S) = 0.03+x

So, Probability of the trucks having both defects is given by

[tex]P(S\cup B)=P(S)+P(B)-P(S\cap B)\\\\0.13=0.03+x+0.06+x-x\\\\0.13-0.03-0.06=2x-x\\\\0.04=x[/tex]

So, Probability of trucks having both defects is 4%.

The percentage of trucks with both defects is 0%.

To solve this problem, we will use the principle of inclusion-exclusion. Let's assume that the percentage of trucks with both defects is x. The percentage of trucks with only defective steering is 3%, the percentage of trucks with only defective brake linings is 6%, and the percentage of trucks with neither defect is 87%. We want to find the percentage of trucks with both defects.

According to the principle of inclusion-exclusion, the total percentage of trucks with at least one defect is given by:

Total = Percentage with defective steering + Percentage with defective brake linings - Percentage with both defects

Using the given percentages, we can set up the equation:

100% = 3% + 6% - x

Simplifying the equation, we get:

x = 9% - 100% = -91%

Since a negative percentage does not make sense in this context, there are no trucks with both defects.

Answer:

The root of the equation [tex]x^3-0.2589x^{2}+0.02262x-0.001122=0[/tex] is x ≈ 0.162035

Step-by-step explanation:

To find the roots of the equation [tex]x^3-0.2589x^{2}+0.02262x-0.001122=0[/tex] you can use the Newton-Raphson method.

It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess [tex]x_{0}[/tex] for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

This is the expression that we need to use

[tex]x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}[/tex]

For the information given:

[tex]f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262[/tex]

For the initial value [tex]x_{0}[/tex] you can choose [tex]x_{0}=0[/tex] although you can choose any value that you want.

So for approximation [tex]x_{1}[/tex]

[tex]x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021[/tex]

Next, with [tex]x_{1}=0.0496021[/tex] you put it into the equation

[tex]f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150[/tex], you can see that this value is close to 0 but we need to refine our solution.

For approximation [tex]x_{2}[/tex]

[tex]x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883[/tex]

Again we put [tex]x_{2}=0.168883[/tex] into the equation

[tex]f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307[/tex] this value is close to 0 but again we need to refine our solution.

We can summarize this process in the following table.

The approximation [tex]x_{5}[/tex] gives you the root of the equation.

When you plot the equation you find that only have one real root and is approximate to the value found.