1. sqrt 52 = 7.211 rounds to 7
2. irrational number
3. sqrt 441 = 21....rational
4. 12^2 + 3^2 = h^2
144 + 9 = h^2
153 = h^2
sqrt 153 = h
12.4 = h <===
5. 6^2 + b^2 = 18^2
36 + b^2 = 324
b^2 = 324 - 36
b^2 = 288
b = sqrt 288
b = 16.97 rounds to 17 <==
6. 39^2 + 52^2 = c^2
1521 + 2704 = c^2
4225 = c^2
sqrt 4225 = c
65 = c
(39 + 52) - 65 = 91 - 65 = 26 miles shorter <==
7. 4^2 + b^2 = 16^2
16 + b^2 = 256
b^2 = 256 - 16
b^2 = 240
b = sqrt 240
b = 15.49 rounds to 15.5 <==
8. ?
To solve this problem, we simply use the equation of
volume for hollow cylinders:
V = π h (R^2 – r^2)
where V is volume, h is height = 10, R is outer radius =
8.4/2 = 4.2, r is inner radius = 6/2 = 3
V = π (10) (4.2^2 – 3^2)
<span>V = 86.4π = 271.43 mm^3</span>
These are the same line when you graph. So there are infinite solutions to this system.
Answer: Choice D.
Max: f (-1,-2)=4; min:f(3,5)=-11
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Work Shown:
Plug in (x,y) = (-1,3)
f(x,y) = -2x-y
f(-1,3) = -2*(-1)-3
f(-1,3) = 2-3
f(-1,3) = -1
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Plug in (x,y) = (3,5)
f(x,y) = -2x-y
f(3,5) = -2*3-5
f(3,5) = -6-5
f(3,5) = -11
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Plug in (x,y) = (4,-1)
f(x,y) = -2x-y
f(4,-1) = -2*4-(-1)
f(4,-1) = -8+1
f(4,-1) = -7
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Plug in (x,y) = (-1,-2)
f(x,y) = -2x-y
f(-1,-2) = -2*(-1)-(-2)
f(-1,-2) = 2+2
f(-1,-2) = 4
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The four outputs are: -1, -11, -7, and 4
The largest output is 4 and that happens when (x,y) = (-1,-2)
So the max is f(x,y) = 4
The smallest output is -11 and that happens when (x,y) = (3,5)
So the min is f(x,y) = -11
This all points to choice D being the answer.
Answer:
(1, 2 )
Step-by-step explanation:
Given the 2 equations
- 2x - y = - 4 → (1)
x + 2y = 5 → (2)
Multiplying (2) by 2 and adding result to (1) will eliminate x- term
2x + 4y = 10 → (3)
Add (1) and (3) term by term to eliminate x
0 + 3y = 6
3y = 6 ( divide both sides by 3 )
y = 2
Substitute y = 2 into either of the 2 equations and solve for x
Substituting into (2)
x + 2(2) = 5
x + 4 = 5 ( subtract 4 from both sides )
x = 1
solution is (1, 2 )