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3 years ago

11

sam goes to the market to buy a hamburger for cookout. he buys 3 3/4 pounds if a hamburger. how many 1/4 pound patties can he ma

ke

1 answer:

Alexandra [31]3 years ago

7 0

3 3/4÷ 1/4
15/4÷1/4
15/4÷4/1
15*4=60
4*1=4
60/4=15
15 times 1/4 can fit into 3 3/4.

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The weight of meteorite A is 5 times the weight of meteorite B. If the sum of their weights is 126 tons, find the weight of each

Sholpan [36]

Answer:

Step-by-step explanation:

if you divide 126 by 5, you get Meteorite B, 25.2, then minus that from 132, and you get Meteorite A, 106.8

To check, do 25.2 x 5 and you should get 132.

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2 years ago

Which value or in the solution set of the inequality − 2/3 x + 13 ≥ − 1

Svetach [21]

Answer:

x $\leq$ 21

Step-by-step explanation:

(-2/3)x+13≥ -1

(-2/3)x≥ -14 (I subtracted by 13 on both sides)

x $\leq$ -14*(-3/2) (I divided by -2/3 on both sides)

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2 years ago

Help I need to get this done I have 15 questions

ivanzaharov [21]

Answer:

8 cm

Step-by-step explanation:

(See the image attached for details)

The blue segment measures 15 cm
The green segment measures 7 cm
The green and yellow segment when added form the blue segment
So: ? + 7 = 15

Solve:

? + 7 = 15

Subtract 7 on both sides:

? + 7 = 15

-7 -7

? = 8 cm

Therefore, the yellow segment, or the missing side length, measures 8 cm.

4 0

3 years ago

If Elisa gets an allowance of $35 per week and spends $10 of it on food and drink, $5 on entertainment and $5 on miscellaneous t

Maslowich

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15$

Step-by-step explanation:

35-10=25$

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6 0

3 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

8 0

2 years ago