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4 years ago

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you are a computer-aided design technican preparing a scale drawing of an office buliding. the scale of the drawing is 1inch=4 f

eet. the front of the building will be 120 feet long. what is the length of the building front on your drawing

1 answer:

Klio2033 [76]4 years ago

5 0

Divide 4 by 1, you get 4. Divide 120 by 4, get 30. Hope this helped!

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What is the domain of this function

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The domain is the set from which the function "starts", i.e. it is the set of the inputs you feed the function with.

In this case, the function starts from the set $\{-2, 2, 3, 4\}$ and lands on the set $\{3, 9, 12\}$

So, the domain is the set $\{-2, 2, 3, 4\}$ .

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3 years ago

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Select all of the terms that apply to the shape. pentagon hexagon heptagon octagon nonagon decagon regular polygon irregular pol

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Answer: Convex, heptagon, regular polygon

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6w-4z+2y+8x=52<br> 2y-4w-6x-2z=-28<br> 6z+2x-4w+4y=36<br> 4x+4y-2w-2z=20<br> solve for z

Nadusha1986 [10]

Answer:

Step-by-step explanation:

1)w=−4/3x+−1/3y+2/3z+26/3

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4 years ago

Match each differential equation to a function which is a solution.

inessss [21]

Answer: 1 - C

2 - E

3 - no answer

4 - B

Step-by-step explanation:

A. $y = 3x+x^2$

$y' = 3 + 2x\\y'' = 2$

Replace in 1:

$y'' + y = 0$

$2 + 3x + x^2 \neq 0$

So, A is not an answer for 1

Replace in 2:

$y' = 3y$

$3+2x = 3(3x + x^2)$

So, A is not an answer for 2

Replace in 3

$2x^2y'' + 3xy'= y$

$2x^2(2) + 3x(3 + 2x) = 4x^2 + 9x + 6x^2 \neq 3x+x^2$

So, A is not an answer for 3

Replace in 4

$y'' + 6y' + 8y = 0$

$2 + 6(3+2x)+8(3x + x^2) = 2+18+12x+24x+8x^2 \neq 0$

So, A is not an answer for 4

B. $y = e^{-4x}$

$y' = -4e^{-4x}$

$y'' = 16e^{-4x}$

Replace in 1

$y'' + y = 0$

$16e^{-4x} -4e^{-4x} = 12e^{-4x} \neq 0$

So, B is not an answer for 1

Replace in 2

$y' = 3y$

$-4e^{-4x} \neq 3e^{-4x}$

So, B is not an answer for 2

Replace in 3

$2x^2y'' + 3xy' = y$

$2x^2(16e^{-4x}) +3x(-4e^{-4x}) \neq e^{-4x}$

So, B is not an answer for 3

Replace in 4

$y'' + 6y' +8y = 0$

$16e^{-4x} + 6(-4e^{-4x}) + 8e^{-4x} = e^{-4x}(16-24+8) = 0$

So, B is an answer for 4

C. $y = sin(x)$

$y' = cos(x)$

$y'' = -sin(x)$

Replace in 1

$y'' + y = 0$

$-sin(x) + sin(x) = 0$

So, C is an answer for 1

We jump to

D. $y = x^{12}$

$y' = 12x^{11}$

$y'' = 132 x^{10}$

Replace in 2

$y' = 3y$

$12x^{11} \neq 3x^{12}$

So, D is not an answer for 2

Replace in 3

$2x^2y'' + 3xy' = y$

$2x^2(132x^{10}) + 3x(12x^{11}) = 300x^{12} \neq x^{12}$

So, D is not an answer for 3

E. $y = 6e^{3x}$

$y' = 18e^{3x}$

$y'' = 54e^{3x}$

Replace in 2

$y' = 3y$

$18e^{3x} = 3(6e^{3x}) = 18e^{3x}$

So, E is an answer for 2

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3 years ago

Given the points (–3,k) and (2,0), for which values of k would the distance between the points be √34 ?

myrzilka [38]

Answer:

The values of k are 3 and -3

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$(-3,k)\\(2,0)$

$d=\sqrt{34}$

substitute

$\sqrt{34}=\sqrt{(0-k)^{2}+(2+3)^{2}}$

$\sqrt{34}=\sqrt{(-k)^{2}+(5)^{2}}$

$\sqrt{34}=\sqrt{k^{2}+25}$

squared both sides

$34=k^{2}+25$

$k^{2}=34-25$

$k^{2}=9$

$k=(+/-)3$

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3 years ago

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