A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $3,000 or $6,000. If the partnership raised $258,000, then how many investors contributed $3,000 and how many contributed $6,000?

x = $3,000 investors
y =
$6,000 investors

Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- and y-variables. Solve the system by the elimination method. Be sure to state your final answer in terms of the original question.

A jar contains 70 nickels and dimes worth $5.70. How many of each kind of coin are in the jar?

Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- and y-variables. Solve the system by the elimination method. Be sure to state your final answer in terms of the original question.

The concession stand at an ice hockey rink had receipts of $7400 from selling a total of 3000 sodas and hot dogs. If each soda sold for $2 and each hot dog sold for $3, how many of each were sold?

x= soda

y= hotdogs

Answer:

2500

Step-by-step explanation:

We have to find the largest product of two numbers whose sum is 100.

Let the two numbers be x and y.

Thus, we can write x+y=100

We can calculate the value of y as:

y = 100 - x

The product of these number can be written as: (x)(y) = (x)(100-x) = 100x - x²

Let f(x) = 100x - x²

Now, the first derivative of this function with respect to x is

[tex]\frac{df(x)}{dx}[/tex] = 100-2x

Equating [tex]\frac{df(x)}{dx}[/tex] = 0, we get,

100-2x = 0

⇒ x = 50

Now, we find the second derivative of the the function f(x) with respect to x

[tex]\frac{d^2f(x)}{dx^2}[/tex] = -2

Since, [tex]\frac{d^2f(x)}{dx^2}[/tex] < 0, then by double derivative test the function have a local maxima at x = 50

This, x = 50 and y = 100-50 =50

Largest product = (50)(50) = 2500

The probability that all six cells are able to replicate is approximately 0.0051, while the probability that at least one cell is not capable of replication is approximately 0.9949.

To solve this problem, we need to use the concept of probability and combinations.

There are 37 - 12 = 25 cells capable of replication. Out of these, we need to select 6 cells. The probability of selecting a cell capable of replication is 25/37 for the first selection, multiplied by 24/36 for the second selection, and so on, until 20/32 for the sixth selection. So, the probability is:

P(all 6 cells able to replicate) = (25/37) * (24/36) * (23/35) * (22/34) * (21/33) * (20/32) ≈ 0.0051

The probability that at least one cell is not capable of replication is equal to 1 minus the probability that all six cells are able to replicate. So, the probability is:

P(at least one cell not able to replicate) = 1 - P(all 6 cells able to replicate) ≈ 0.9949

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Answer:

1) yes

2) no, (fog)(x)=3-x/2

3) no, (f o g)(x)= -2x+6/3x-5

The first and second pair of functions are not inverses of each other, while the third pair of functions are inverses.

The first pair of functions, f(x) = 2x + 3/(4x - 3) and g(x) = 3x + 3/(4x - 2), are not inverses of each other. To determine this algebraically, we need to calculate (f o g)(x) and (g o f)(x) and check if they equal to x. In this case, (f o g)(x) is x + 2/3, which is not equal to x, therefore f and g are not inverses of each other.

The second pair of functions, f(x) = -x^3 + 2 and g(x) = 3(cubedroot)x - 2/(2), are also not inverses. By calculating (f o g)(x) and (g o f)(x), we found that (f o g)(x) = x - 14/8, which is not equal to x.

The third pair of functions, f(x) = -2x + 4/(2 - 5x) and g(x) = (4 - 2x)/(5 - 2x), is indeed inverses of each other. By calculating (f o g)(x) and (g o f)(x), we found that (f o g)(x) = x and (g o f)(x) = x, which means f and g are inverses of each other.

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Answer:

the rate is 0.208

Answer:

Quality A is greater

Step-by-step explanation:

Answer:

7/10

Step-by-step explanation:

The total number of marbles is 7 + 2 + 1 = 10.  So the probability of selecting a brown marble is 7/10

Answer:

[tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex]

Step-by-step explanation:

We have been given that adhesive tape made from fabric has a tensile strength of not less than 20.41 kg/2.54 cm of width. We are asked to reduce these quantities to grams and millimeters.

We know 1 kg equals 1000 grams and 1 cm equals 10 mm.

[tex]\frac{20.41\text{ kg}}{\text{2.54 cm}}[/tex]

[tex]\frac{20.41\text{ kg}}{\text{2.54 cm}}\times \frac{\text{1 cm}}{\text{10 mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{2.54\times\text{10 mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{254\text{ mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{254\text{ mm}}\times \frac{\text{1000 grams}}{\text{1 kg}}[/tex]

[tex]\frac{20.41\times \text{1000 grams}}{254\text{ mm}}[/tex]

[tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex]

Therefore, our required measurement would be [tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex].

Answer:

[tex]y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

[tex]y"+y'-2y=1[/tex]

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

[tex]D^2+D-2\ =\ 0[/tex]

[tex]=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}[/tex]

[tex]=>\ D=\ -2\ or\ 1[/tex]

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

[tex]y_c(t)\ =\ C_1e^{-2t}+C_2e^t[/tex]

Let's assume that

     [tex]y_1(t)=e^{-2t}[/tex]          [tex]y_2(t)=e^t[/tex]

[tex]=>\ y'_1(t)=-2e^{-2t}[/tex]        [tex]y'_2(t)=e^t[/tex]

   and g(t)=1

Now, the Wronskian can be given by

[tex]W=y_1(t).y'_2(t)-y'_1(t).y_2(t)[/tex]

   [tex]=e^{-2t}.e^t-e^t(-e^{-2t})[/tex]

   [tex]=e^{-t}+2e^{-t}[/tex]

   [tex]=3e^{-t}[/tex]

Now, the particular solution can be given by

[tex]y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}[/tex]

[tex]=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}[/tex]

[tex]=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}[/tex]

[tex]=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}[/tex]

[tex]=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]

Now, the complete solution of the given differential equation can be given by

[tex]y(t)\ =\ y_c(t)+y_p(t)[/tex]

      [tex]=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]

Answer:

The linear problem is to maximize [tex]Z = C_ {1} X_ {1} + C_ {2}X_ {2} = 60X_ {1} + 50X_ {2}[/tex], s.a.

subject to

[tex]\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\\\\\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\\\\X_ {2} \geq 2\\\\X_ {1}, X_ {2} \geq 0[/tex]

Step-by-step explanation:

Let the decision variables be:

[tex] X_ {1} [/tex]: number of units of product 1 to produce.

[tex] X_ {2} [/tex]: number of units of product 2 to produce.

Let the contributions be:

[tex]C_ {1} = 60\\\\C_ {2} = 50[/tex]

The objective function is:

[tex]Z = C_{1} X_{1}+ C_{2}X_{2} = 60X_ {1} + 50X_ {2}[/tex]

The restrictions are:

[tex]\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\\\\\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\\\\X_ {2} \geq 2\\\\X_ {1}, X{2} \geq 2\\\\[/tex]

The linear problem is to maximize [tex]Z = C_ {1} X_ {1} + C_ {2}X_ {2} = 60X_ {1} + 50X_ {2}[/tex], s.a.

subject to

[tex]\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\\\\\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\\\\X_ {2} \geq 2\\\\X_ {1}, X_ {2} \geq 0[/tex]

Answer:

The value of x is [tex]\frac{(3c+b)}{3-c}[/tex].

Step-by-step explanation:

The given equation is

[tex]cx+b=3(x-c)[/tex]

Using distributive property we get

[tex]cx+b=3(x)+3(-c)[/tex]

[tex]cx+b=3x-3c[/tex]

To solve the above equation isolate variable terms.

Subtract 3x and b from both sides.

[tex]cx-3x=-3c-b[/tex]

Taking out common factors.

[tex]x(c-3)=-(3c+b)[/tex]

Divide both sides by (c-3).

[tex]x=-\frac{(3c+b)}{c-3}[/tex]

[tex]x=\frac{(3c+b)}{3-c}[/tex]

Therefore the value of x is [tex]\frac{(3c+b)}{3-c}[/tex].

Answer:

The statement is true for every n between 0 and 77 and it is false for [tex]n\geq 78[/tex]

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: [tex]\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4[/tex]

For n=1: [tex]\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4[/tex]

From this point we will assume that [tex]n\geq 2[/tex]

As we can see, [tex]\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4[/tex] and [tex](4n)^4=256n^4[/tex]. Then,

[tex]\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4[/tex]

Now, we will use the formula for the sum of the first 4th powers:

[tex]\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}[/tex]

Therefore:

[tex]\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0[/tex]

and, because [tex]n \geq 0[/tex],

[tex]465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1[/tex]

Observe that, because [tex]n \geq 2[/tex] and is an integer,

[tex]n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5[/tex]

In concusion, the statement is true if and only if n is a non negative integer such that [tex]n\leq 77[/tex]

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  [tex](4n)^4- \sum^{n}_{i=0} (2i)^4[/tex] for 77 and 78 you will obtain:

[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=53810064[/tex]

[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992[/tex]

We will use mathematical induction to prove this inequality.

We will prove this inequality using mathematical induction. First, let's check the base case when n = 0. The left-hand side (LHS) of the inequality is 0 and the right-hand side (RHS) is (4*0)^4 = 0. So, the inequality holds for n = 0.

Next, assume that the inequality holds for some positive integer k, i.e.,

∑([2i]^4) ≤ (4k)^4 (where the sum is taken from i = 0 to k)

We will prove that it also holds for k + 1. Adding the (k+1)th term to both sides of the inequality:

∑([2i]^4) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)

Now, simplifying the LHS and RHS:

(∑([2i]^4)) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)

Answer:

The first two coins are quarters, and the one on the right is a nickle.

the two quarters [0.25+0.25] is 0.50 cents. Add the nickle [0.5] and you have 0.55 cents!

ex:     0.25+0.25+0.5

             0.50+0.5

                 =0.55 (cents)

Step-by-step explanation:

Answer:

Step-by-step explanation:

Sample size of 8 infants were taken and their birth lengths in inches recorded and also 6 months lengths.

If x is length at birth time, and y 6 month length

we have as per table below.

x y x^2 y^2 xy

1 19.75 25.5 390.0625 650.25 503.625

2 20.5 26.25 420.25 689.0625 538.125

3 19 25 361 625 475

4 21 26.75 441 715.5625 561.75

5 19 25.75 361 663.0625 489.25

6 18.5 25.25 342.25 637.5625 467.125

7 20.25 27 410.0625 729 546.75

8 20 26.5 400 702.25 530

Total 158 208 3125.625 5411.75 4111.625

[tex]∑ x,     ∑ x2 , ∑ y,           ∑ xy,           ∑ y2\\158 208 3125.625 5411.75 4111.625[/tex]

SSxx = 3125.625 and SSyy = 5411.75

Answer:

[tex]f'(0)=0[/tex]

Step-by-step explanation:

Applying the chain rule

[tex]\frac{d}{dx} (f(-x))=-\frac{df}{dx}[/tex]

Then it becomes

[tex]\frac{df}{dx} =-\frac{df}{dx}[/tex]

In x=0

[tex]\frac{d[tex]f'(0)=-f'(0)\\f'(0)+f'(0)=0\\2f'(0)=0\\[/tex]f}{dx} =-\frac{df}{dx}[/tex]

Then

[tex]f'(0)=0[/tex]

Answer:

There is a 47.50% probability that the chosen senator is a Democrat.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula:

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In your problem we have that:

A(what happened) is the probability of a gun owner being chosen:

There are 100 people in the survay(53 Democrats, 45 Republicans ans 2 Independents), and 40 of them have guns(19 Democrats, 21 Republicans). So, the probability of a gun owner being chosen is:

[tex]P(A) = \frac{40}{100} = 0.4[/tex]

[tex]P(A/B)[/tex] is the probability of a senator owning a gun, given that he is a Democrat. 19 of 53 Democrats own guns, so the probability of a democrat owning a gun is:

[tex]P(A/B) = \frac{19}{53} = 0.3585[/tex]

[tex]P(B)[/tex] is the probability that the chosen senators is a Democrat. There are 100 total senators, 53 of which are Democrats, so:

[tex]P(B) = \frac{53}{100} = 0.53[/tex]

If a senator participating in that survey was picked at random and turned out to be a gun owner, what was the probability that he or she was a Democrat?

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.53)*(0.3585)}{(0.40)} = 0.4750[/tex]

There is a 47.50% probability that the chosen senator is a Democrat.

Answer:

110,000 base 2

Step-by-step explanation:

column 1 [the first position in the number]:

1+1=0, (carry 1)

column 2:

0+1 +1 (carried)=0, (carry 1)

column 3:

1+0+1 (carried)=0, (carry 1)

column 4:

0+1+1 (carried)=0, (carry 1)

column 5:

1+1+1=1, (carry 1)

then you write the last 1 'cause there is n number to add with:

[tex]10,101_{2}+11,011_{2}=110,000_{2}[/tex]

In binary system the highest number to write is 1, if you add 1+1, it jumps to 0, and you have to carry 1 to the next position.

If you are not sure about the sum, you can convert the numbers in base 2, to base 10, so you can know if it is correct:

[tex]10,101_{2}=21_{10}\\11,011_{2}=27_{10}\\110,000_{2}=48_{10}[/tex]

So 21+27=48.

In decimal system when you add 9+1, it jumps to 0 and then you have to carry 1 to the next position, because the the highest number you can write is 9.

Answer:

The answer is no: the set of all 2x2 matrices such that det(A)=0 is not a subspace of the vector space of all 2x2 matrices.

Step-by-step explanation:

In order for a set of matrices to be a subspace of all 2x2 matrices, three conditions must be satisfied:

1) That the set is not empty.

2) If A and B are both 2x2 matrices with zero determinant then the matrix A+B should also be a matrix with zero determinant.

3) The determinant of c*A, where "c" is any complex number and A is any matrix of the set, should be zero.

1)

The first condition is satisfied by the set of all 2x2 matrices such that det(A)=0, since there are plenty 2x2 matrices with zero determinant.

2)

The second condition is not satisfied, since from the determinant properties, we know that:

[tex]det(A+B)\geq det(a)+det(B)[/tex]

The equality might hold, but it is not a general characteristic. For example, if we consider the following matrices:

[tex]A = \left[\begin{array}{cc}1&0\\0&0\end{array}\right], \quad B = \left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]

We can easily check that the determinant of both matrices is zero, nevertheless the determinant of the sum is different than zero.

Therefore, the set of all 2x2 matrices such that det(A)=0 is not a subspace of the vector space of all 2x2 matrices.

The set of all 2x2 matrices such that det(A) = 0 is not a subspace of the vector space of all 2x2 matrices.

To determine if a set is a subspace of a vector space, it must satisfy three conditions:

1. The set must contain the zero vector (in this case, the zero matrix).

2. The set must be closed under addition, meaning that if A and B are matrices in the set, then A + B must also be in the set.

3. The set must be closed under scalar multiplication, meaning that if A is a matrix in the set and c is a scalar, then cA must also be in the set.

Let's examine each condition:

1. The zero matrix, denoted by O, has a determinant of det(O) = 0, so it is included in the set. This condition is satisfied.

2. Consider two 2x2 matrices A and B with determinant 0:

[tex]\[ A = \begin{pmatrix} a b \\ c d \end{pmatrix}, \quad \text{det}(A) = ad - bc = 0 \][/tex]

 [tex]\[ B = \begin{pmatrix} e f \\ g h \end{pmatrix}, \quad \text{det}(B) = eh - fg = 0 \][/tex]

 The sum of A and B is:

 [tex]\[ A + B = \begin{pmatrix} a+e b+f \\ c+g d+h \end{pmatrix} \][/tex]

 The determinant of A + B is:

 [tex]\[ \text{det}(A + B) = (a+e)(d+h) - (b+f)(c+g) \][/tex]

 [tex]\[ \text{det}(A + B) = ad + ah + ea + eh - bc - bg - cf - fg \][/tex]

 Since det(A) = 0 and det(B) = 0, we have:

 [tex]\[ ad - bc = 0 \][/tex]

 [tex]\[ eh - fg = 0 \][/tex]

 However, this does not imply that det(A + B) = 0. The cross terms ah, ea, bg, and cf may not sum to zero, and thus det(A + B) may not be zero. Therefore, the set is not closed under addition, and this condition is not satisfied.

3. Consider a scalar c and a matrix A with determinant 0:

 [tex]\[ cA = c \begin{pmatrix} a b \\ c d \end{pmatrix} = \begin{pmatrix} ca cb \\ cc cd \end{pmatrix} \][/tex]

 The determinant of cA is:

 [tex]\[ \text{det}(cA) = (ca)(cd) - (cb)(cc) = c^2(ad - bc) \][/tex]

 Since det(A) = 0, we have ad - bc = 0, and thus det[tex](cA) = c^2(0) = 0[/tex]for any scalar c. This means that the set is closed under scalar multiplication, and this condition is satisfied.

Answer:

[tex]\text{Density}=\frac{0.556\text{ Grams}}{\text{ cm}^3}[/tex]

Step-by-step explanation:

We are asked to find the density of a cylinder whose volume is 1000.0 cubic cm and mass is 556 grams.

[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]

Substitute the given values:

[tex]\text{Density}=\frac{556\text{ Grams}}{1000.0\text{ cm}^3}[/tex]

[tex]\text{Density}=\frac{0.556\text{ Grams}}{\text{ cm}^3}[/tex]

Therefore, the density of the metallic cylinder is 0.556 grams per cubic centimeter.

Answer:

Solution: [tex]B=5+25e^{4-4t}[/tex]

Step-by-step explanation:

Given: [tex]\dfrac{dB}{dt}+4B=20[/tex]  

with B(1)=30

The differential equation in form of linear differential equation,

[tex]\dfrac{dy}{dt}+Py=Q[/tex]

Integral factor, IF: [tex]e^{\int Pdt}[/tex]

General Solution:

[tex]y\cdot IF=\int Q\cdot IFdt[/tex]

[tex]\dfrac{dB}{dt}+4B=20[/tex]  

P=4, Q=20

IF= [tex]e^{\int 4dt}=e^{4t}[/tex]

Solution:

[tex]Be^{4t}=\int 20e^{4t}dt[/tex]

[tex]Be^{4t}=5e^{4t}+C[/tex]

[tex]B=5+Ce^{-4t}[/tex]

B(1)=30 , Put t=1, B=30

[tex]30=5+Ce^{-4}[/tex]

[tex]C=25e^4[/tex]

[tex]B=5+25e^{4-4t}[/tex]

Answer:

Pigs = 20 and Chickens = 50

Step-by-step explanation:

Let the number of pigs be x

and let the number of chicken be y

Thus, x + y = 70

Since Chicken has 2 legs and pigs has 4 legs.

⇒ 4x + 2y = 180

Solving both equations,

We get, x = 20 and y = 50

Thus number of pigs = 20

and, number of chickens = 50.

Answer:

The solution of the system of linear equations is [tex]x=3, y=4, z=1[/tex]

Step-by-step explanation:

We have the system of linear equations:

[tex]2x+3y-6z=12\\x-2y+3z=-2\\3x+y=13[/tex]

Gauss-Jordan elimination method is the process of performing row operations to transform any matrix into reduced row-echelon form.

The first step is to transform the system of linear equations into the matrix form. A system of linear equations can be represented in matrix form (Ax=b) using a coefficient matrix (A), a variable matrix (x), and a constant matrix(b).

From the system of linear equations that we have, the coefficient matrix is

[tex]\left[\begin{array}{ccc}2&3&-6\\1&-2&3\\3&1&0\end{array}\right][/tex]

the variable matrix is

[tex]\left[\begin{array}{c}x&y&z\end{array}\right][/tex]

and the constant matrix is

[tex]\left[\begin{array}{c}12&-2&13\end{array}\right][/tex]

We also need the augmented matrix, this matrix is the result of joining the columns of the coefficient matrix and the constant matrix divided by a vertical bar, so

[tex]\left[\begin{array}{ccc|c}2&3&-6&12\\1&-2&3&-2\\3&1&0&13\end{array}\right][/tex]

To transform the augmented matrix to reduced row-echelon form we need to follow these row operations:

[tex]\left[\begin{array}{ccc|c}1&3/2&-3&6\\1&-2&3&-2\\3&1&0&13\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\3&1&0&13\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\0&-7/2&9&-5\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&-7/2&9&-5\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&3&3\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&1&1\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&0&4\\0&0&1&1\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&3/2&0&9\\0&1&0&4\\0&0&1&1\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|c}1&0&0&3\\0&1&0&4\\0&0&1&1\end{array}\right][/tex]

From the reduced row echelon form we have that

[tex]x=3\\y=4\\z=1[/tex]

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution.

Answer:

5

Step-by-step explanation:

-7x-3x+2= -8x-8;

-10x+2= -8x-8;

-10x+2+8x= -8;

-10x+8x= -8-2;

-2x= -10;

x=(-10)/(-2);

x=5.

Answer:

Ans. the annual rate of return, in order to turn $5,000 into $24,000 in 15 years is 11.02% annual.

Step-by-step explanation:

Hi, well, in order to find the value of the interest rate of return, we need to solve for "r" the following equation,

[tex]Future Value=PresentValue(1+r)^{n}[/tex]

Where:

n= years (time that the money was invested)

r=annual rate of return (Decimal)

So, let´s see the math of this.

[tex]24,000=5,000(1+r)^{15}[/tex]

[tex]\frac{24,000}{5,000} =(1+r)^{15}[/tex]

[tex]\sqrt[15]{\frac{24,000}{5,000} } =1+r[/tex]

[tex]\sqrt[15]{\frac{24,000}{5,000} } -1=r[/tex]

[tex]r=0.11023[/tex]

So the annual rate of return that turns $5,000 into $24,000 in 15 years is 11.02%.

N=15; PV=5,000; FV=24,000; PMT=N.A; I/Y=11.02% P/Y=N.A

Best of Luck.

The problem is a step-by-step calculation with several possible combinations of type A and type B vehicles when a total of 101 counts are registered. A systematic approach is required to find all possible whole number solutions.

This problem is an example of a diophantine problem or a linear equation in two variables. If we denote the number of type A vehicles by 'a', and the number of type B vehicles by 'b', the problem can be represented by the equation 2a + 9b = 101.

As you are looking for all possible solutions, you have to do a systematic search. You will find that:

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Answer:

x=2

Step-by-step explanation:

−3(4−x)+3x=3(10−5x)

(−3)(4)+(−3)(−x)+3x=(3)(10)+(3)(−5x)

−12+3x+3x=30+−15x

(3x+3x)+(−12)=−15x+30

6x+−12=−15x+30

6x−12=−15x+30

6x−12+15x=−15x+30+15x

21x−12=30

Step 3: Add 12 to both sides.

21x−12+12=30+12

21x=42

Answer:

-½

Step-by-step explanation:

Simplify \frac{25}{100}

100

25

to \frac{1}{4}

4

1

.

-\sqrt{\frac{1}{4}}−√

4

1

2 Simplify \sqrt{\frac{1}{4}}√

4

1

to \frac{\sqrt{1}}{\sqrt{4}}

√

4

√

1

.

-\frac{\sqrt{1}}{\sqrt{4}}−

√

4

√

1

3 Simplify \sqrt{1}√

1

to 11.

-\frac{1}{\sqrt{4}}−

√

4

1

4 Since 2\times 2=42×2=4, the square root of 44 is 22.

-\frac{1}{2}−

2

1

Done

Decimal Form: -0.5

Answer: D.)

Step-by-step explanation:

The first step would be to reduce the fraction in the square root, so divide both the numerator and denominator by 25 to get 1/4.

Then calculate the root, any root of one equals one so that stays as is. The exponential form of 4 would be 2^2.

Then reduce the index of the radical and exponent with 2 and you get answer D

The sum of 7.25 L and 875 cL, reduced to milliliters, is 16,000 mL as per the concept of addition.

To add 7.25 L and 875 cL, we need to convert the centiliters to liters before performing the addition.

1. Convert 875 cL to liters:

Since there are 100 centiliters in a liter, we divide 875 by 100 to get the equivalent in liters:

875 cL ÷ 100 = 8.75 L

2. Now that both quantities are in liters, we can add them together:

7.25 L + 8.75 L = 16 L

3. Finally, to convert the result to milliliters, we multiply by 1000 since there are 1000 milliliters in a liter:

16 L × 1000 = 16,000 mL

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Final answer:

To add 7.25 L and 875 cL and reduce the result to milliliters, convert 875 cL to liters to get 8.75 L, then add it to 7.25 L to total 16 L. Finally, convert 16 L to milliliters by multiplying by 1,000, resulting in 16,000 mL.

Explanation:

To add 7.25 L and 875 cL and reduce the result to milliliters, you first need to understand the conversion between liters, centiliters, and milliliters. Remember, there are 1000 milliliters (mL) in a liter (L) and 10 milliliters in a centiliter (cL).

First, let's convert 875 cL to liters to simplify the addition. Since there are 10 mL in a cL, and 1000 mL in a L, you would convert as follows:

875 cL = 875 / 10 = 87.5 mL

However, we need to recognize the proper conversion to liters in the step above. Correctly, it should state: 875 cL = 8.75 L (since 100 cL = 1 L).

Once we have both measurements in liters, we can easily add them:

7.25 L + 8.75 L = 16.0 L

To convert the total liters to milliliters, multiply by 1,000 (since there are 1,000 mL in 1 L).

16.0 L × 1,000 = 16,000 mL

Answer:

Step-by-step explanation:

Given that the solution of a certain differential equation is of the form

[tex]y(t) = ae^{7t} +be^{11t}[/tex]

Use the initial conditions

i) y(0) =1

[tex]1=a(1)+b(1)\\a+b=1[/tex] ... I

ii) y'(0) = 4

Find derivative of y first and then substitute

[tex]y'(t) = 7ae^{7t} +11be^{11t}\\y'(0) =7a+11b \\7a+11b =4 ...II[/tex]

Now using I and II we solve for a and b

Substitute b = 1-a in II

[tex]7a+11(1-a) = 4\\-4a+11 =4\\-4a =-7\\a = 1.75 \\b = -0.75[/tex]

Hence solution is

[tex]y(t) = 1.75e^{7t} -0.75e^{11t}[/tex]

Answer:

y(t) = a exp(3t) + b exp(4t) conditions, y(0) = 3 y'(0) = 3 y(0) = a exp(3 x 0) + b exp(4 x 0) = a exp(0) + b exp(0) = (a x 1) + (b x 1) = a + b y'(0) = 0 so the linear equation is, a + b = 3

Step-by-step explanation:

During the flood season, the water rises 26.4 feet up the side of the lake.

The angle of elevation is given as 15 degrees 30 minutes,

Now, it can be converted to decimal degrees as 15.5 degrees.

Let's denote the distance up the side of the lake as x feet.

Now set up a trigonometric equation using the tangent function:

tan(15.5°) = (7.3 feet) / x

We can solve for x by rearranging the equation:

x = (7.3 feet) / tan(15.5°)

Evaluating this expression gives us:

x = (7.3 feet) / 0.277

x = 26.4 feet

Therefore, the water rises 26.4 feet up the side of the lake.

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