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Strike441 [17]
2 years ago
12

What is the value of h(2)? -2 -1 0 3

Mathematics
1 answer:
Nesterboy [21]2 years ago
4 0

Answer:

0

Step-by-step explanation:

all the question is asking is

if x is 2 then what is y

the line on the left is downward facing line which ends at

(2,0) <- where is x is 2 and y is 0

the upward facing line on the right does start at (2,3)

but the dot there is a

hollow dot

hollow dot represents excluded value

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A lidless box is to be made using 2m^2 of cardboard find the dimensions of the box that requires the least amount of cardboard
Jlenok [28]
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3 . Find the dimensions of the box that requires the least amount of cardboard. Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From these, we obtain x 2y = 8 = xy2 . This forces x = y = 2, which forces z = 1. Calculating second derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus, moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
5 0
3 years ago
345 degrees to a radion​
miv72 [106K]

We convert degrees to radian by multiplying by  (

\frac{\pi}{180})

=  (345)  ×  (  \frac{\pi π}{180} )

=  (\frac{345}{180})  ×  (π)

=  1.9166π

=  1.9166  ×  3.14

=  6.018 rad

≈  6.02 rad

4 0
3 years ago
Find the surface area of the right triangular prism shown ​
Inessa05 [86]

The surface area of the triangular prism is 144 cm²

<u>Explanation:</u>

Given:

Base of the triangle = 4

Height of the triangle = 3

Hypotenuse of the triangle = 5

Length of the prism = 11

Surface area of the rectangular prism = ?

Surface area, A = ah+bh+ch+\frac{1}{2} \sqrt{-a^4 + 2(ab)^2 + 2(ac)^2 - b^4 + 2(bc)^2 - c^4}

On substituting the value, we get:

A = 4 X 11+ 3X11+5X11+\frac{1}{2} \sqrt{-(4)^2 + 2(4X3)^2 + 2(4X5)^2 - 3^4 + 2(3X5)^2-5^4} \\\\A = 144 cm^2

Therefore, the surface area of the triangular prism is 144 cm²

6 0
3 years ago
The functions f and g are defined by these sets if input and output values.
erastovalidia [21]
(x,y)
x=input
y=output
example
we see
g=(1,2)
theefor
g(1)=2


a.
f(4)=1
g(1)=2
g(f(4))=2

b.
g(-2)=4
f(4)=1
f(g(-2))=1

c.
f(3)=5
g(5)=5
f(5)=0
f(g(f(3)))=0
5 0
3 years ago
Te which theorem you can use to show that the quadrilate
lukranit [14]

Answer:

Step-by-step explanation

c

6 0
2 years ago
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