Suppose that Jane borrows $8400 from a bank on April 9 at an annual rate of 9.2% simple interest. How much does she owe on August 20 of the same year?

Answer:

P(B) = 0.30

Step-by-step explanation:

This is a probability problem that can be modeled by a diagram of Venn.

We have the following probabilities:

[tex]P(A) = P_{A} + P(A \cap B) = 0.50[/tex]

In which [tex]P_{A}[/tex] is the probability that only A happens.

[tex]P(B) = P_{B} + P(A \cap B) = P_{B} + 0.15[/tex]

To find P(B), first we have to find [tex]P_{B}[/tex], that is the probability that only B happens.

Finding [tex]P_{B}[/tex]:

The problem states that P(A OR B) = 0.65. This is the probability that at least one of this events happening. Mathematically, it means that:

[tex]1) P_{A} + P(A \cap B) + P_{B} = 0.65[/tex]

The problem states that P(A) = 0.5 and [tex]P(A \cap B) = 0.15[/tex]. So we can find [tex]P_{A}[/tex].

[tex]P(A) = P_{A} + P(A \cap B)[/tex]

[tex]0.5 = P_{A} + 0.15[/tex]

[tex]P_{A} = 0.35[/tex]

Replacing it in equation 1)

[tex]P_{A} + P(A \cap B) + P_{B} = 0.65[/tex]

[tex]0.35 + 0.15 + P_{B} = 0.65[/tex]

[tex]P_{B} = 0.65 - 0.35 - 0.15[/tex]

[tex]P_{B} = 0.15[/tex]

Since

[tex]P(B) = P_{B} + P(A \cap B)[/tex]

[tex]P(B) = 0.15 + 0.15[/tex]

[tex]P(B) = 0.30[/tex]

Answer:

A 1.4% of the total connectors are expected to fail during the warranty period.

Step-by-step explanation:

Let's assume a population of 1000 connectors (to make the math easiest) and let's analize the dry connectors.  

Of the 1000 connectors, 900 are kept dry. and of that number, 9 are the ones that fails during the warranty period.  (90% of 1000 is 900. 1% of 900 is 9)

Of the 1000 connectors, 100 are wet. And of that number, 5 are connectors that will fail during the warranty period. (10% of 1000 is 100, and 5% of 100 is 5)

So overall we have 14 connectors that will fail from 1000 connectors.

That is a 1.4% of the total samples.

Answer:

Age of Alice=25 years

Age of Jane=20 years

Step-by-step explanation:

We are given that the age of Jane is 80 % of the age Alice.

We have to find the age of Jane and Alice.

Let x be the age of Alice

According to question

Age of Jane=80% of Alice=80% of x=[tex]\frac{80}{100}\times x=\frac{4x}{5}[/tex]

[tex]x+\frac{4x}{5}=45[/tex]

[tex]\frac{5x+4x}{5}=45[/tex]

[tex]\frac{9x}{5}=45[/tex]

[tex]x=\frac{45\times 5}{9}=25[/tex]

Age of Alice=25 years

Age of Jane=[tex]\frac{4}{5}\times 25=20 years[/tex]

Age of Jane=20 years

Over the 28-day cycle of taking Triphasil-28 birth control tablets, a total of 1.925 mg of levonorgestrel and 0.680 mg of ethinyl estradiol are consumed.

The Triphasil-28 birth control tablets are taken in a sequence of 21 active pills containing hormones and 7 inactive pills (placebos). To calculate the total amount of levonorgestrel and ethinyl estradiol consumed over the 28-day period, each phase needs to be taken into account.

In phase 1, there are 6 tablets each with 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol. So, 6 tablets * 0.050 mg = 0.300 mg of levonorgestrel and 6 tablets * 0.030 mg = 0.180 mg of ethinyl estradiol.

In phase 2, there are 5 tablets each with 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol. So, 5 tablets * 0.075 mg = 0.375 mg of levonorgestrel and 5 tablets * 0.040 mg = 0.200 mg of ethinyl estradiol.

In phase 3, there are 10 tablets each with 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol. So, 10 tablets * 0.125 mg = 1.250 mg of levonorgestrel and 10 tablets * 0.030 mg = 0.300 mg of ethinyl estradiol.

Adding up all the quantities, we have a total of 0.300 mg + 0.375 mg + 1.250 mg = 1.925 mg of levonorgestrel and 0.180 mg + 0.200 mg + 0.300 mg = 0.680 mg of ethinyl estradiol consumed over the 21 active days of the 28-day cycle.

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Answer:

35 people did no travel by any of those modes of transportation.

Step-by-step explanation:

Reviewing the groups described, we can see there are 170 that traveled by either plane or train and 50 traveled by bus, but not plane or train.

This means that the 50 people that traveled by bus are not included in the group of 170 that used either plane or train.

Since those two groups include people that used bus (50 people) and people that used either plane or train (170 people), the union of these two sets include all the people that used at least one of the tree modes of transportation mentioned (bus, plane or train).

The result of the union between these two sets is 50+170=220 people which travel by at least one of those modes of transportation.

Since the total number of surveyed adults is 255, the number of people that did not travel by any of these modes of transportations is 255-220=35 people.

Answer:

[tex]\frac{7}{6}[/tex]

Step-by-step explanation:

[tex]-4\frac{1}{2} + 5\frac{2}{3} =[/tex]

For the first term you have to multiply 2x(-4) and add 1, for the second term you have to multiply 3x5 and add 2.

[tex]\frac{2(-4)+1}{2} + \frac{3(5)+2}{3} =[/tex]

[tex]-\frac{9}{2} +\frac{17}{3} =[/tex]

Now you need to find the lowest common multiple between the denominators, just cross multiply as it is shown:

[tex]-\frac{9}{2} (\frac{3}{3} )+\frac{17}{3} (\frac{2}{2} )=[/tex]

[tex]-\frac{27}{6} +\frac{34}{6} = \frac{7}{6}[/tex]

finally you get the result by doing a substraction = 7/6, or 1[tex]\frac{1}{6}[/tex]

Answer:

The solutions of the equation are 0 and 0.75.

Step-by-step explanation:

Given : Equation [tex]16x^4 - 24x^3 +9x^2 =0[/tex]

To find : All solutions of the equation algebraically. Use a graphing utility to verify the solutions graphically ?

Solution :

Equation [tex]16x^4 - 24x^3 +9x^2 =0[/tex]

[tex]x^2(16x^2-24x+9)=0[/tex]

Either [tex]x^2=0[/tex] or [tex]16x^2-24x+9=0[/tex]

When [tex]x^2=0[/tex]

[tex]x=0[/tex]

When [tex]16x^2-24x+9=0[/tex]

Solve by quadratic formula, [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

[tex]x=\frac{-(-24)\pm\sqrt{(-24)^2-4(16)(9)}}{2(16)}[/tex]

[tex]x=\frac{24\pm\sqrt{0}}{32}[/tex]

[tex]x=\frac{24}{32}[/tex]

[tex]x=\frac{3}{4}[/tex]

[tex]x=0.75[/tex]

The solutions of the equation are 0 and 0.75.

For verification,

In the graph where the curve cut x-axis is the solution of the equation.

Refer the attached figure below.

In logic, own symbols are used in order to be able to represent the relations between propositions in a general and independent way to the proposition, in order to be able to find the relationship process that operates in the communicated message, the propositional logic.

For this purpose there are, among others, the following logical operators: conjunction (and) ∧, disjunction (or) ∨, denial (not) ¬,  conditional (if - then) ⇒ and double conditional (if and only if, iff) ⇔.

So for this case we have:

Answer

a) Sam had pizza last night if and only if Chris finished her homework.

p⇔q

b) Pat watched the news this morning iff Sam did not have pizza last night.

r⇔¬p

c) Pat watched the news this morning if and only if Chris finished her homework and Sam did not have pizza last night.

r⇔(q∧¬p)

d) In order for Pat to watch the news this morning, it is necessary and sufficient that Sam had pizza last night and Chris finished her homework.

r⇔(p∧q)

e) q ⇔ r

Chris finished his homework if and only if Pat watched the news this morning

f) p ⇔ (q ∧ r)

Sam had pizza last night if and only if Chris finished his homework and Pat watched the news this morning

g) (¬p) ⇔ (q ∨ r)

Sam didn't have pizza last night if and only if Chris finished his homework or Pat watched the news this morning

h) r ⇔ (p ∨ q)

Pat watched the news this morning if Sam had pizza last night or Chris finished his homework

Answer:

The system will have no solution when [tex]k = -4[/tex] and [tex]h \neq 7.5[/tex].

Step-by-step explanation:

We can find these values by the Gauss-Jordan Elimination method.

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

[tex]-3x - 3y = h[/tex]

[tex]-4x + ky = 10[/tex]

This system has the following augmented matrix:

[tex]\left[\begin{array}{ccc}-3&-3&h\\-4&k&10\end{array}\right][/tex]

The first thing i am going to do is, to help the row reducing:

[tex]L_{1} = -\frac{L_{1}}{3}[/tex]

Now we have

[tex]\left[\begin{array}{ccc}1&1&-\frac{h}{3}\\-4&k&10\end{array}\right][/tex]

Now I want to reduce the first row, so I do:

[tex]L_{2} = L_{2} + 4L_{1}[/tex]

So:

[tex]\left[\begin{array}{ccc}1&1&-\frac{h}{3}\\0&k+4&10 - \frac{4h}{3}\end{array}\right][/tex]

From the second line, we have

[tex](k+4)y = 10- \frac{4h}{3}[/tex]

The system will have no solution when there is a value dividing 0, so, there are two conditions:

[tex]k+4 = 0[/tex] and [tex]10 - \frac{4h}{3} \neq 0[/tex]

[tex]k+4 = 0[/tex]

[tex]k = -4[/tex]

...

[tex]10 - \frac{4h}{3} \neq 0[/tex]

[tex]\frac{4h}{3} \neq 10[/tex]

[tex]4h \neq 30[/tex]

[tex]h \neq \frac{30}{4}[/tex]

[tex]h \neq 7.5[/tex]

The system will have no solution when [tex]k = -4[/tex] and [tex]h \neq 7.5[/tex].

Answer:

a) There are 11,881,376 of these words in total.

b) 7,893,600 of these words contain no repeated letters.

c) 896,376 of these words start with an a or end with a z or both.

Step-by-step explanation:

The English alphabet has 26 letters.

Our word has the following format

C1 - C2 - C3 - C4 - C5

C1 is the first character, C2 the second, etc...

(a) How many of these words are there total?

Each of C1, C2,... can be 26. So the total is

[tex]T = (26)^{5} = 11,881,376[/tex]

There are 11,881,376 of these words in total.

(b) How many of these words contain no repeated letters?

C1 can be any letter. C2 can be any letter bar the letter at C1. C3 any other than C2, C1... So

26-25-24-23-22

[tex]T = 26*25*24*23*22 = 7,893,600[/tex]

7,893,600 of these words contain no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

[tex]T = T_{1} + T_{2} + T_{3}[/tex]

[tex]T_{1}:[/tex] Start with a, end with any letter other than z.

1-26-26-26-25

[tex]T_{1} = 26^{3}*25 = 439,400[/tex]

[tex]T_{2}:[/tex]End with z, start with any other letter than A

25-26-26-26-1

[tex]T_{2} = 26^{3}*25 = 439,400[/tex]

[tex]T_{3}:[/tex]Start with A, end with Z

1-26-26-26-1

[tex]T_{3} = 17,576[/tex]

[tex]T = T_{1} + T_{2} + T_{3} = 439,400 + 439,400 + 17,576 = 896,376[/tex]

896,376 of these words start with an a or end with a z or both.

Answer:

%82.5

Step-by-step explanation:

From point 1 we know that Moe´s grade just before taking the final exam represents 60% of the final grade. Then, using the information in the point 2 we can compute Moe´s final grade as follows:

[tex]FG=0.40*FE+0.60*0.45[/tex],

where FG is Moe´s Final Grade and FE is Moe´s final exam grade. Then,

[tex]\frac{ FG-0.60*0.45}{0.40}=FE[/tex].

So, in order to receive the passing grade average of 60% for the class Moe needs to obtain in his exam:

[tex]FE=\frac{ 0.60-0.60*0.45}{0.40}=0.825[/tex]

That is, he need al least %82.5 to obtain a passing grade.

Answer:  The proof is done below.

Step-by-step explanation:  Given that A is a 3 x 3 matrix such that det (A) = 9.

We are to prove the following :

[tex]det(3A^{-1})=3.[/tex]

For a non-singular matrix B of order n, we have two two properties of its determinant :

[tex](i)~det(B^{-1})=\dfrac{1}{det(B)},\\\\\\(ii)~det(kB)=k^ndet(B),~\textup{k is a scalar.}[/tex]

Therefore, we get

[tex]det(A^{-1})=\dfrac{1}{det(A)}=\dfrac{1}{9},[/tex]

and so,

[tex]det(3A^{-1})~~~~~~~[\textup{since A is of order 3}]\\\\=3^3det(A^{-1})\\\\=27\times\dfrac{1}{9}\\\\=3.[/tex]

Hence proved.

Answer:

(a) Set A = {1,2,3}

    Set B = {1,2,4,5}

(b) Set A Ç B = {1,2}

     Set A e B = {1,2}

Step-by-step explanation:

As per the question,

Given data :

A ∪ B = read as A union B = {1,2,3,4,5}

A ∩ B = read as A intersection B =  {1,2}

(a) An example of sets A and B such that A ∩ B = {1,2} and A U B = {1,2,3,4,5).

So, first draw the Venn-diagram, From below Venn diagram, One of the possibility for set A and set B is:

Set A = {1,2,3}

Set B = {1,2,4,5}

(b) An example of sets A and B such that A Ç B and A e B,

Set A Ç B read as common elements of set A in Set B.

Therefore,

Set A Ç B = {1,2}

Set A e B implies that which element/elements of A is/are present in set B.

Therefore,

Set A e B = {1,2}

Answer:

one

Step-by-step explanation:

I think, there were 4 lines and the two lines are reaching each other

Answer:

4:3

Step-by-step explanation:

You count up all the red marbles which equals 4 and put them on one side, then you add up all the rest of the marbles same color or not which equals 3 and put it on the other side of the 4

If a  physician prescribes 2.5 million units of penicillin G potassium daily for 1 week. To provide the prescribed dosage regimen of 2.5 million units of penicillin G potassium daily for 1 week we would need 42 tablets, each containing 250 mg.

What is the Number of tablets needed ?

Total amount of penicillin G potassium required

Since there are 7 days in a week the total amount required is:

Total amount required =2.5 million units/day * 7 days

Total amount required = 17.5 million units

Convert the units of penicillin G potassium to micrograms (mcg)

Total amount required in micrograms:

Total amount required = 17.5 million units * 0.6 mcg/unit

Total amount required = 10.5 million mcg

Convert micrograms to milligrams (mg)

Convert the total amount required to milligrams:

10.5 million mcg / 1000

= 10,500 mg

The number of tablets needed:

Number of tablets needed:

Number of tablets needed = 10,500 mg / 250 mg/tablet

Number of tablets needed  = 42 tablets

Therefore to provide the prescribed dosage regimen of 2.5 million units of penicillin G potassium daily for 1 week we would need 42 tablets, each containing 250 mg.

The patient needs to take 42 tablets of penicillin G potassium over one week. This is determined by converting the prescribed units to milligrams and dividing by the tablet dosage.

Calculating Dosage of Penicillin G Potassium

To determine how many tablets of penicillin G potassium are needed, we need to go through several steps of unit conversion.

Therefore, the patient needs to take 42 tablets over the course of one week.

Answer:

1015 widgets

Step-by-step explanation:

First investigate how many widgets can machine A produce in 1 hour:

If it produces 435 in 3 hours, then it will produce one third of that in one hour:

In ONE (1) hour [tex]\frac{435}{3} = 145[/tex] widgets

Therefore in seven (7) hours it will produce seven times the amount it does in one hour, that is:

[tex]145 * 7 = 1015[/tex] widgets

Answer:

answer is 1015.

Step-by-step explanation:

Answer:

[tex]-4x+y=-5[/tex]

Step-by-step explanation:

It is given that he vector y = ai + bj is perpendicular to the line ax + by = c.

Given given point is P(-1,-9) and given vector is v = -4i + j.

Here, a = -4 and b= 1.

We need to find the equation of the line through P perpendicular to v.

Using the above fact, the vector v = -4i + j is perpendicular to the line

[tex](-4)x+(1)y=c[/tex]

[tex]-4x+y=c[/tex]          ... (1)

The line passes through the point (-1,-9).

Substitute x=-1 and y=-9 in the above equation to find the value of c.

[tex]-4(-1)+(-9)=c[/tex]

[tex]4-9=c[/tex]

[tex]-5=c[/tex]

The value of c is -5.

Substitute c=-5 in equation (1).

[tex]-4x+y=-5[/tex]

Therefore the equation of line which passes through P and perpendicular to v is -4x+y=-5.

Answer: 49 people

Step-by-step explanation:

First you need to separate the people that read both from the people that read the daily news. Do this by subtracting 171 - 40 = 131.

Now you can subtract the total amount of people that only read the Daily Mews from the total amount of people polled (198-131=67)

From here you need to subtract the amount of people that read neither from the remaining total. (67-18=49)

this is where you get the 49 people out of 198 that read the Sun Gazette

The number of people who read the Sun Gazette is calculated by subtracting the number of people who only read the Daily News and those who read neither from the total polled. The result is 49 Sun Gazette readers.

This problem involves the mathematics concept of set theory, specifically union and intersection of sets. Here, the total number of people polled are considered as the 'Universal Set'. The 'Daily News readers' and the 'Sun Gazette readers' are the two subsets of this Universal set.

The data given can be interpreted as follows:

Since 40 people read both, they are being counted twice in the 171 (Daily News readers) figure. Hence, we subtract 40 from 171.

The number of people who only read the Daily News = 171 - 40 = 131

We subtract this number and those who read neither from the total polled to find the number of Sun Gazette readers.

The number of Sun Gazette readers = 198 - 131 (only Daily News readers) - 18 (neither) = 49 Sun Gazette readers

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The domain of the function represented in the graph is 30 SXS 60 or 30 to 60. Hence option C is correct.

Given that,

Five-year-old students at an elementary school were involved.

They were given a 30-yard head start in a race.

The graph represents the distance run by the average student in 30 seconds.

The graph is related to the age of the runner.

Since we can see that,

The graph starts from 30 yards and ends at 60 yards

Therefore,

The domain of the function represented in the graph is 30 SXS 60 or 30 to 60, which is option C.

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Answer:

[tex]f''(x)=f''(-x)[/tex]

Step-by-step explanation:

A function satisfying the equation [tex]f(x)=f(-x)[/tex] is said to be an even function. This denomination comes from the fact that the same relation is satisfied for functions of the form [tex]x^{n}[/tex] with [tex]n[/tex] even. Observe that if [tex]f[/tex] is twice differentiable we can derivate using the chaing rule as follows:

[tex]f(x)=f(-x)[/tex] implies [tex]f'(x)=f'(-x)\cdot (-1)=-f'(-x)[/tex]

Applying the chain rule again we have:

[tex]f'(x)=-f'(-x)[/tex] implies [tex]f''(x)=-f''(-x)\cdot (-1)=f''(-x)[/tex]

So we have that function [tex]f''(x)[/tex] is also an even function.

Step-by-step explanation:

.454 kilograms= 1 pound.

Multiply .454 by 160

4.54

160

--------

000

2724 0<--Place marker

45400<--Double Place Marker

- - - - - - - -

72640 <-- Add

To find decimal point, count decimal place (Number of digits after the decimal on both numbers you multiply together) In this case, it's 2 ( 5 and 4 in 4.54) So, you count two spaces from right to left in your answer and tah dah!

72.640 (Zero isn't needed- just a placemarker)

Hope I was helpful :)

Answer:

5-card hands with at least one club: [tex] {52 \choose 5}-{39 ]choose 5}[/tex]

5-card hands with at least two cards of the same rank: [tex]{52 \choose 5}-{13 \choose 5}4^5[/tex]

Step-by-step explanation:

To determine how many 5-card hands have at least one club, we can count how many do NOT have at least one club, and then subtract that from the total amount of 5-card hands that there are.

A 5-card hand that doesn't have at least one club, is one whose 5 cards are from spades,hearts or diamonds. Since a standard deck of cards has 13 clubs, 39 of the cards are spades, hearts of diamonds. Getting a 5-card hand out of those cards, is choosing 5 cards out of those 39 cards. So there are [tex] {39 \choose 5}[/tex] 5-card hands without any clubs.

The total amount of 5-card hands is [tex]{52 \choose 5}[/tex], since a 5-card hand is simply a group of 5 cards out of the full deck, which has 52 cards.

Therefore the number of 5-card hands that have at least one club is [tex] {52 \choose 5}-{39 \choose 5}[/tex].

To determine how many 5-card hands have at least two cards with the same rank we can follow the same approach. We determine how many 5-card hands have NO cards with the same rank, and the subtract that out of the total amount of 5-card hands.

A 5-card hand that doesn't have cards of the same rank, is a group of 5 cards all from different ranks. Such hand can be made then by choosing first which 5 different ranks are going to be present in the hand, out of the 13 available ranks. So there are [tex] {13 \choose 5}[/tex] possible combinations of ranks. Then, choosing which card from each of the chosen ranks is the one that is going to be in the hand, is choosing which of 4 cards from EACH rank is going to be in the hand. So for each rank there are 4 availble choices, and so there are [tex]4^5[/tex] possible ways to choose the specific cards from each rank that will be in the hand. So the amount of 5-card hands with all ranks different is [tex] {13 \choose 5}\cdot{4^5}[/tex]

Therefore the amount of 5-card hands with at least two cards with the same rank is  [tex]{52 \choose 5}-{13 \choose 5}\cdot4^5[/tex]

Answer:

b

Step-by-step explanation:

Answer with Step-by-step explanation:

We are given that two functions f and g are decreasing function for all real numbers .

We have to prove that fog is increasing.

Increasing Function: If [tex]x_1 \leq x_2[/tex]

Then, [tex]f(x_1)\leq f(x_2)[/tex]

Decreasing function: if [tex]x_1\leq x_2[/tex]

Then, [tex]f(x_1)\geq f(x_2)[/tex]

Suppose [tex]f(x)=-x[/tex]

[tex]g(x)=-2x[/tex]

fog(x)=f(g(x))=f(-2x)=-(-2x)=2x

Therefore, f and g are decreasing function for all real numbers  then fog is  increasing.

Hence, proved

Answer:

(a) [tex]\frac{dy}{(2y+1)}=tdt[/tex] (b) [tex]y=\frac{e^{t^2}+e^{2c}-1}{2}[/tex]

Step-by-step explanation:

(1) We have given [tex]\frac{dy}{dt}+ycost=0[/tex]

[tex]\frac{dy}{dt}=-ycost[/tex]

[tex]\frac{dy}{y}=-costdt[/tex]

Integrating both side

[tex]lny=-sint+c[/tex]

[tex]y=e^{-sint}+e^{-c}[/tex]

(2) [tex]\frac{dy}{dt}-2ty=t[/tex]

[tex]\frac{dy}{dt}=2ty+t[/tex]

[tex]\frac{dy}{dt}=t(2y+1)[/tex]

[tex]\frac{dy}{(2y+1)}=tdt[/tex]

On integrating both side  

[tex]\frac{ln(2y+1)}{2}=\frac{t^2}{2}+c[/tex]

[tex]ln(2y+1)={t^2}+2c[/tex]

[tex]2y+1=e^{t^2}+e^{2c}[/tex]

[tex]y=\frac{e^{t^2}+e^{2c}-1}{2}[/tex]

Answer with Step-by-step explanation:

We are given that [tex]v_1,v_2,..,v_4[/tex] are in [tex]R^4[/tex] and [tex]v_1,v_2,v_3[/tex] is linearly dependent then {v_1,v_2,v_3,v_4}[/tex] is also linearly dependent.

We have to find that given statement is true or false.

Dependent vectors:Dependent vectors are those vectors in which atleast one vector is  a linear combination of other given vectors.

Or If we have vectors [tex]x_1,x_2,....x_n[/tex]

Then their linear combination

[tex]a_1x_1+a_2x_2+.....+a_nx_n=0[/tex]

There exist at least one scalar which is not zero.

If [tex]v_1,v_2,v_3[/tex] are dependent vectors then

[tex]a_1v_1+a_2v_2+a_3v_3=0[/tex] for scalars [tex]a_1,a_2,a_3[/tex]

Then , by definition of dependent  vectors

There exist a vector which is not equal to zero

If vector [tex]v_3[/tex] is a linear combination of [tex]v_1\;and \;v_2[/tex], So at least one of vectors in the set is a linear combination of others and the set is linearly dependent.

Hence, by definition of dependent vectors

{[tex]v_1,v_2,v_3,v_4[/tex]} is linearly dependent.

Option B is true.

Answer:

360 inch

Step-by-step explanation:

Step 1: Converting

6inch = 6inch

5ft = 60inch

Step 2: Finding the area

Area of a rectangle = L x B

A = 6 x 60

A = 360inch

Answer:

Step-by-step explanation:

There are six basic types of quadrilaterals:

Then we have:

Quadrilaterals with NOT ALWAYS congruent diagonals:

Quadrilaterals with ALWAYS congruent diagonals:

Answer:

Step-by-step explanation:

Write expressions for the number of students in each major. Then write the equation needed to relate them the way the problem statement says they are related.

For year y, the number of students in each major is ...

1) Engineering is twice Business:

  120 +22y = 2(105 -4y) . . . . . Engineering is double Business in year y

  120 +22y = 210 -8y . . . . . . . eliminate parentheses

  120 +30y = 210  . . . . . . . . . . add 8y

  4 + y = 7 . . . . . . . . . . . . . . . . divide by 30

  y = 3  . . . . . . . . subtract 4

In 3 years there will be 2 times as many students majoring in Engineering than in Business.

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2) Engineering is twice Biology:

  120 +22y = 2(98 +6y) . . . . Engineering is double Biology in year y

  120 +22y = 196 +12y . . . . . eliminate parentheses

  120 +10y = 196 . . . . . . . . . . subtract 12y

  10y = 76 . . . . . . . . . . . . . . . .subtract 120

  y = 7.6 . . . . . . divide by 10

In 8 years there will be 2 times as many students majoring in Engineering than in Biology.