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Natasha_Volkova [10]
3 years ago
12

Please help with these math questions part one

Mathematics
1 answer:
katrin [286]3 years ago
3 0
V of cone=(1/3)(pi)(h)(r^2)
V of cylinder=(pi)(h)(r^2)


1.
 r=6
h=10
v=(1/3)(pi)(10)(6^2)
v=1/3pi10(36)
v=1/3pi360
v=120pi
answer is B

2.
r=d/2
d=10
10/2=5=r
 V=(3.14)(5)(5^2)
V=3.14(125)
V=392.5
answer is B

3. 9.2, 3.7
V=(pi)(9.2)(3.7^2)
V=(pi)(9.2)(13.69)
V=(pi)(125.948)
input 125.948

4. v=(1/3)(pi)(12)(3^2)
V=(1/3)(pi)(12)(9)
V=(pi)(12)(3)
V=pi(36
v=36pi
v=36(3.14)
V=113.04

input 113.04


ANSERS
1.B
2.B
3. 129.948
4. 113.04
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Mike wants to fence in part of his backyard. He wants the length of the fenced-in area to be at least 20 feet long, l ≥ 20. He h
zhuklara [117]

I added a screenshot of the complete question along with the given choices.

<em><u>Answers:</u></em>

w = 10 ft and l = 50 ft ................> second option

w = 20 ft and l = 60 ft ...............> third option

w = 50 ft and l = 40 ft ...............> fifth option

<em><u>Explanation:</u></em>

<u>We are given that:</u>

length should be at least 20 ...........> length ≥ 20

equation for perimeter is : 2l + 2w ≤ 200

<u>We will check each option as follows:</u>

<u>Option 1:</u>

w = 50 ft and l = 10 ft

This option is incorrect as the proposed length is less than 20.

<u>Option 2:</u>

w = 10 ft and l = 50 ft

Since the length is greater than 20, the first condition is satisfied.

Now, we check the second condition:

2l + 2w = 2(50) + 2(10) = 100 + 20 = 120 ≤ 200

The second condition is also satisfied.

This option is correct

<u>Option 3:</u>

w = 20 ft and l = 60 ft

Since the length is greater than 20, the first condition is satisfied.

Now, we check the second condition:

2l + 2w = 2(60) + 2(20) = 120 + 40 = 160 ≤ 200

The second condition is also satisfied.

This option is correct

<u>Option 4:</u>

w = 90 ft and l = 30 ft

Since the length is greater than 20, the first condition is satisfied.

Now, we check the second condition:

2l + 2w = 2(30) + 2(90) = 60 + 180 = 240 > 200

The second condition is not satisfied.

This option is incorrect

<u>Option 5:</u>

w = 50 ft and l = 40 ft

Since the length is greater than 20, the first condition is satisfied.

Now, we check the second condition:

2l + 2w = 2(50) + 2(40) = 100 + 80 = 180 ≤ 200

The second condition is also satisfied.

This option is correct

Hope this helps :)

8 0
3 years ago
Read 2 more answers
Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu
kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

f(x) =  {x}^{5}  -  {x}^{3}  + 6

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b) Using calculus, we find the first derivative of the given function.

f'(x) = 5 {x}^{4} - 3 {x}^{2}

At turning point f'(x)=0.

5 {x}^{4} - 3 {x}^{2}  = 0

This implies that,

{x}^{2} (5 {x}^{2}  - 3) = 0

{x}^{2}  = 0 \: or \: 5 {x}^{2}  - 3 = 0

x =  - \frac{ \sqrt{15} }{5}   \: or \: x = 0 \:  \: or \: x =\frac{ \sqrt{15} }{5}

We plug this values into the original function to obtain the y-values of the turning points

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We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

f''(x) = 20 {x}^{3}  - 6x

f''( -  \frac{ \sqrt{15} }{5} ) \:   <  \: 0

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(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  + 750))

is a maximum point.

f''( \frac{ \sqrt{15} }{5} ) \:    >  \: 0

Hence

(     \frac{ \sqrt{15} }{5}  , \frac{1}{125} (- 6 \sqrt{15}  + 750))

is a minimum point.

f''(0) \: =\: 0

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4 0
3 years ago
X+ 0.0825×=216.50 A.) x=2.00 B.) x=20.0 C.) x=200 D.) x=2000
Luden [163]
X+0.0825x=216.5
1.0825x=216.5
x=216.5 / 1.0825
x=200

Answer: C.)  x=200
8 0
3 years ago
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