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stepan [7]
3 years ago
9

The base area of a square-based prism is 25 ft squared. One of the lateral faces of the prism is 30 ft squared. What is the surf

ace area of the prism?
Mathematics
1 answer:
telo118 [61]3 years ago
7 0

Answer:

  • 170 ft²

Step-by-step explanation:

Surface area = 4*lateral faces + 2*bases

  • A = 4*30 + 2*25 = 170 ft²
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Step-by-step explanation:

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Please resolve this problem for me! I've been having trouble with this for a while.
Bad White [126]

Answer:

a) 41

b) -31

c) 25/3

Step-by-step explanation:

The instructions explain what a function is pretty clearly.

Basically, whatever is in the parentheses next to the f, you plug in to the expression that is the function.

ex:     f(x) = x + 4

        f(5) = 5 + 4 = 9

Okay, onto the problems:

a) f(x) = 3x - 1

First, plug in 14, since that is given:

f(14) = 3(14) - 1

and solve.

f(14) = 42 - 1

f(14) = 41

So, 41 is the answer to a.

b) Do the same thing you did in problem a. Plug in the number inside the parentheses and solve.

f(-10) = 3(-10) - 1

f(-10) = -30 - 1

f(-10) = -31

-31 is the answer to b.

c) For this problem, you have to do a bit of algebra.

First, take the base function f(x):

f(x) = 3x - 1

Then, set f(x) = 24:

24 = 3x - 1

and solve by isolating x.

25 = 3x

\dfrac{25}{3} = x

x = \dfrac{25}{3}

25/3 is the answer to c.

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x−2y+3z=8, 3y+z=12, −2x+2z=−4 Enter your answer, in the form (x,y,z), in the boxes in simplest terms.
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How to find left and right inverse of a 3x2 matrix.
Mnenie [13.5K]

Let A be a 3×2 matrix, L its left inverse, and R its right inverse. L and R are then matrices such that LA = I₂ (the 2×2 identity matrix) and AR = I₃ (the 3×3 identity matrix). Clearly L must be 2×3 and R must be 3×2 in order for the matrix products to be defined.

To find L and R, we start by introducing a square matrix on the the left sides of either equation above. In particular, we uniformly multiply both sides by the transpose of A, then solve for the inverse.

For the left inverse, we have

LA=I

(LA)A^\top = IA^\top

L\left(AA^\top\right) = A^\top

\left(L\left(AA^\top\right)\right)\left(AA^\top\right)^{-1} = A^\top \left(AA^\top\right)^{-1}

L\left(\left(AA^\top\right)\left(AA^\top\right)^{-1}\right) = A^\top \left(AA^\top\right)^{-1}

LI = A^\top \left(AA^\top\right)^{-1}

L = A^\top \left(AA^\top\right)^{-1}

We do the same thing for the right inverse, but take care with how we multiply both sides of AR = I₃.

AR=I

A^\top(AR)=A^\top I

\left(A^\top A\right)R = A^\top

\left(A^\top A\right)^{-1} \left(\left(A^\top A\right)R\right) = \left(A^\top A\right)^{-1} A^\top

\left(\left(A^\top A\right)^{-1} \left(A^\top A\right)\right) R = \left(A^\top A\right)^{-1} A^\top

IR = \left(A^\top A\right)^{-1} A^\top

R = \left(A^\top A\right)^{-1} A^\top

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2 years ago
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