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Darina [25.2K]
3 years ago
15

Teresa needs to save $25.50 to go to the movies this weekend. She currently has $3. If Teresa makes $7.50 an hour babysitting, h

ow many hours does she need to babysit this week?
Mathematics
2 answers:
victus00 [196]3 years ago
6 0

Answer: she needs to babysit for 3 hours this week.

Step-by-step explanation:

Let x represent the number of hours that she would babysit in a week.

She currently has $3. If Teresa makes $7.50 an hour babysitting. This means that if she babysits for x hours in a weeks, the total amount of money that she would have saved would be

7.5x + 3

Teresa needs to save $25.50 to go to the movies this weekend. This means that

7.5x + 3 = 25.5

7.5x = 25.5 - 3

7.5x = 22.5

x = 22.5/7.5

x = 3

Aleksandr [31]3 years ago
3 0

If we set the amount of hours she needs to babysit as x, his word problem can be shown as:

3+ 7.50x = 25.50

Now we can just solve for x

25.50-3= 22.50

22.50/7.50 = 3

Therefore she has to babysit 3 hours

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Do the ratios 8:24, and 5:15 form a proportion<br><br> yes<br> no
Dmitrij [34]

Answer:

Yes they do, the proportion of 1:3 as 1 x 8=8, and 3 x 8=24, and 5 x 1=5, and 5 x 3=15

Step-by-step explanation:

5 0
3 years ago
Jorge is asked to build a box in the shape of a rectangular prism. The maximum girth of the box is 20 cm. What is the width of t
MariettaO [177]

Answer:

The width of the box is 6.7 cm

The maximum volume is 148.1 cm³

Step-by-step explanation:

The given parameters of the box Jorge is asked to build are;

The maximum girth of the box = 20 cm

The nature of the sides of the box = 2 square sides and 4 rectangular sides

The side length of square side of the box = w

The length of the rectangular side of the box = l

Therefore, we have;

The girth = 2·w + 2·l = 20 cm

∴ w + l = 20/2 = 10

w + l = 10

l = 10 - w

The volume of the box, V = Area of square side × Length of rectangular side

∴ V = w × w × l = w × w × (10 - w)

V = 10·w² - w³

At the maximum volume, we have;

dV/dw = d(10·w² - w³)/dw = 0

∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0

2×10·w - 3·w² = 20·w - 3·w² = 0

20·w - 3·w² = 0 at the maximum volume

w·(20 - 3·w) = 0

∴ w = 0 or w = 20/3 = 6.\overline 6

Given that 6.\overline 6 > 0, we have;

At the maximum volume, the width of the block, w = 6.\overline 6 cm ≈ 6.7 cm

The maximum volume, V_{max}, is therefore given when w = 6.\overline 6 cm = 20/3 cm  as follows;

V = 10·w² - w³

V_{max} = 10·(20/3)² - (20/3)³ = 4000/27 = 148.\overline {148}

The maximum volume, V_{max} = 148.\overline {148} cm³ ≈ 148.1 cm³

Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)

The width of the box is;

6.7 cm

The maximum volume is;

148.1 cm³

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