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3 years ago

ASAP

Mathematics

1 answer:

lana [24]3 years ago

6 0

Step-by-step explanation:

+-2,-(-2)etc dont know if this is corect tho

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In desperate need of help please :/

Dmitry_Shevchenko [17]

Answer:

since 13 miles is 68640 i wood say 189 inches long from bumper to bumper. A mile is 5,280 feet (or 63,360 inches).

If 335 of these cars were parked with their bumpers touching, they would be almost exactly a mile long.

If your question refers to traffic jams, the answer will be a bit different, since cars in traffic jams don’t get that close together. I’m going to take a wild guess here, but let’s assume that there is a 4-foot gap between cars in our hypothetical traffic jam. now put that in to 13 miles

Step-by-step explanation:

4 0

3 years ago

66.64 rounded to the nearest cent

Zolol [24]

It would just be 66.6

4 0

3 years ago

Explain how place value could be used to find the difference between 4.23 into 2.75

Tomtit [17]

Because without place value you might put the decimal in the wrong place and that will effect the answer big time.

5 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

3 years ago

HELP PLEASEEEEEEEEEeeeee

maks197457 [2]

Might have to experiment a bit to choose the right answer.

In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.

Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456

In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.

Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.

3 0

3 years ago