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Svetlanka [38]
3 years ago
14

Winnie poured 14 cups of water into a rectangular container measuring 13 inches by 7 inches by 6 centimeters. [1 cup = 14.44 cub

ic inches; 1 inch = 2.54 cm] Part A: What is the maximum volume of water that the rectangular container can hold? Show your work. (3 points) Part B: How many cubic inches of water was poured into the container? (3 points) Part C: What was the height of the water in the container? Show your work. (4 points)
Mathematics
1 answer:
Alex_Xolod [135]3 years ago
8 0
Part A: 
<span>Max volume = Volume of container = 13 in x 7 in x 6 in = 546 in^3 </span>

<span>Part B: </span>
<span>1 cup = 14.4375 in^3 </span>
<span>
14 cups = 14.4375 in^3 x 7 = 202.125 in^3 </span>

<span>Part C: </span>
<span>Height of water = (202.125 in^3)/(13 in x 7 in) = 2.22 in.</span>
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Find the derivative of the function by using the Product Rule. Simplify your answer. f(x) = (x - 3)(x + 3)
Morgarella [4.7K]

Answer:

f'(x)=2x

Step-by-step explanation:

Given : Function f(x)=(x-3)(x+3)

To find : The derivative of the function by using the Product Rule ?

Solution :

The product rule of derivative is

\frac{d}{dx}(u\cdot v)=uv'+vu'

Here, u=x-3 and v=x+3

\frac{d}{dx}((x-3)\cdot (x+3))=(x-3)\frac{d}{dx}(x+3)+(x+3)\frac{d}{dx}(x-3)

\frac{d}{dx}((x-3)\cdot (x+3))=(x-3)1+(x+3)1

\frac{d}{dx}((x-3)\cdot (x+3))=x-3+x+3

\frac{d}{dx}((x-3)\cdot (x+3))=2x

Therefore, f'(x)=2x

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3 years ago
Read 2 more answers
(FIXED) Can somebody help me get to x?
Hatshy [7]

18 - 12 = 6. Because the top legs of the 2 triangles are congruent, you divide 6  / 2 to get 3 as their length. Then you can use the Pythagorean Theorem to solve for x.

3^2 + x^2 = 14^2

9 + x^2 = 196

x^2 = 187

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3 years ago
Solve the following congruence equations for X a) 8x = 1(mod 13) b) 8x = 4(mod 13) c) 99x = 5(mod 13)
xxMikexx [17]

Answer:

a) 5+13k  where k is integer

b) 20+13k where k is integer

c)12+13k where k is integer

Step-by-step explanation:

(a)

8x \equiv 1 (mod 13) \text{ means } 8x-1=13k.

8x-1=13k

Subtract 13k on both sides:

8x-13k-1=0

Add 1 on both sides:

8x-13k=1

I'm going to use Euclidean Algorithm.

13=8(1)+5

8=5(1)+3

5=3(1)+2

3=2(1)+1

Now backwards through the equations:

3-2=1

3-(5-3)=1

3-5+3=1

(8-5)-5+(8-5)=1

2(8)-3(5)=1

2(8)-3(13-8)=1

5(8)-3(13)=1

So compare this to:

8x-13k=1

We see that x is 5 while k is 3.

Anyways 5 is a solution or 5+13k is a solution where k is an integer.

b)

8x \equiv 4 (mod 13)

8x-4=13k

Subtract 13k on both sides:

8x-13k-4=0

Add 4 on both sides:

8x-13k=4

We got this from above:

5(8)-3(13)=1

If we multiply both sides by 4 we get:

8(20)-13(12)=4

So x=20 and 20+13k is also a solution where k is an integer.

c)

[tex]99x \equiv 5 (mod 13)[/tex

99x-5=13k

Subtract 13k on both sides:

99x-13k-5=0

Add 5 on both sides:

99x-13k=5

Using Euclidean Algorithm:

99=13(7)+8

13=8(1)+5

Go back through the equations:

13-8=5

13-(99-13(7))=5

8(13)-99=5

99(-1)+8(13)=5

Compare this to 99x-13k=5 and see that x=-1 or -1+13=12 or 12+13k is a solution where k is an integer.

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Answer: 33.3% increase, pretty sure

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