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S_A_V [24]
2 years ago
5

How do you solve this?

Mathematics
1 answer:
Basile [38]2 years ago
6 0

Answer:

NO SOLUTION

Step-by-step explanation:

By the give me brainlist please

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At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawls, each in the same amount. E
spayn [35]

At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawls, each in the same amount.

Let x be  the amount of one of Derek's withdrawals

Each of John's withdrawals was $5 more than each withdrawal that Derek made.

x + 5 is t the amount of one of John's withdrawals

Derek made 7 withdrawals

So amount withdraw 7 times = 7x

John made 5 withdrawals

So amount withdraw 5 times = 5(x+5)

Both Derek and John withdrew the same amount of money in the end

(A) 7x = 5(x+5)

(B) Solve for x

7x = 5x + 25

Subtract 5x from both sides

2x = 25

Divide by 2

x = 12.5

(C) check your solution

we plug in 12.5 for x in 7x= 5x + 25

7(12.5) = 5(12.5) + 25

87.5 = 62.5+ 25

87.5 = 87.5

(D) Each brother withdrawal 87.5 dollars



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3 years ago
What algebraic expression models the word phrase three fewer than the sum of c and d?
Serjik [45]
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A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman
Yanka [14]

Answer:

W\approx 8.72 and L\approx 15.57.

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

2L+W+\frac{1}{2}(2\pi r)=38

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

2L+W+\frac{1}{2}(2r\pi)=38

2L+W+\frac{\pi}{2}W=38

Let us find area of window equation as:

\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)

\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})

\text{Area}=W\cdot L+\frac{\pi}{8}W^2

Now, we will solve for L is terms W from perimeter equation as:

L=38-(W+\frac{\pi }{2}W)

Substitute this value in area equation:

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2

Since we need the area of window to maximize, so we need to optimize area equation.

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2  

A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2  

Let us find derivative of area equation as:

A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W  

A'=38-2W-\pi W+\frac{\pi}{4}W    

A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W

A'=38-2W-\frac{3\pi W}{4}

To find maxima, we will equate first derivative equal to 0 as:

38-2W-\frac{3\pi W}{4}=0

-2W-\frac{3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}*4=-38*4

-8W-3\pi W=-152

8W+3\pi W=152

W(8+3\pi)=152

W=\frac{152}{8+3\pi}

W=8.723210

W\approx 8.72

Upon substituting W=8.723210 in equation L=38-(W+\frac{\pi }{2}W), we will get:

L=38-(8.723210+\frac{\pi }{2}8.723210)

L=38-(8.723210+\frac{8.723210\pi }{2})

L=38-(8.723210+\frac{27.40477245}{2})

L=38-(8.723210+13.70238622)

L=38-(22.42559622)

L=15.57440378

L\approx 15.57

Therefore, the dimensions of the window that will maximize the area would be W\approx 8.72 and L\approx 15.57.

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I hope this helps. :)
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