Ask question

Ask question

All categories

3 years ago

9

Everyday, Sarah drives the same distance to work and back home. If Sarah drove 1,005 miles in 5 days , how far did Sarah drive 3

days?

1 answer:

Aleks04 [339]3 years ago

4 0

The answer is 201 because all you do is divide 1,005 by 5 then to check your answer you multiply 201 by 5 then you get 1,005.

Your welcome

You might be interested in

Factor this expression completely, then place the factors in the proper location on the grid. a^3y + 1

ivolga24 [154]

Answer:

$a^{3y} + 1 = (a^{y}+1 )^{3} - 3a^y(a^{y}+1)\\\\$

Step-by-step explanation:

We are to factorize the expression $a^{3y} + 1$ completely. To do this, we will apply the expression below;

The expression can be rewritten as $a^{3y} + 1^{3}$

To factorize the expression, we need to first factorize $(a^{y}+1 )^{3}$ first

$(a^{y}+1 )^{3} =(a^{y}+1 )(a^{y}+1 )^{2}\\= (a^{y}+1 )((a^y)^{2} } + 2a^{y} +1)\\= (a^y)^{3} +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y+1\\(a^{y}+1 )^{3} = ((a^y)^{3} + 1) +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y\\(a^{y}+1 )^{3} = ((a^y)^{3} + 1) +3(a^y)^{2}+3a^y\\$

The we will make $a^{3y} + 1^{3}$ the subject of the formula as shown;

$(a^y)^{3} + 1 = (a^{y}+1 )^{3} - (3(a^y)^{2}+3a^y)\\(a^y)^{3} + 1^{3} = (a^{y}+1 )^{3} - (3(a^y)^{2}+3a^y)\\\\$

$(a^y)^{3} + 1 = (a^{y}+1 )^{3} - (3(a^y)^{2}+3a^y)\\\\$

$a^{3y} + 1 = (a^{y}+1 )^{3} - (3(a^y)^{2}+3a^y)\\\\$

$a^{3y} + 1 = (a^{y}+1 )^{3} - 3a^y(a^{y}+1)\\\\$

This last result gives the expansion of the expression

7 0

4 years ago

What is the percent of 108 is 81

marusya05 [52]

That would be 133.33

8 0

3 years ago

Read 2 more answers

Triangle RST has vertices R(2, 0), S(4, 0), and T(1, –3). The image of triangle RST after a rotation has vertices

velikii [3]

<span>C) Ro 270 degree (anticlockwise)

============================

</span>

6 0

4 years ago

Read 2 more answers

A washing machine can hold

svet-max [94.6K]

If the washing machine can hold 6 kilograms and he has 4 kilograms of clothes than he would only need to do 1 load

6 0

3 years ago

27. The altitude BD to the base AC of isosceles ABC has a length of 12. If AC = 10,

Vsevolod [243]

The answer is: 36

Because, bisecting the base of the triangle gives of two segments of five. This forms two triangles with base of 5 and altitude (in this case height) of 12. There exists a special pythagorean triplet triangled named 5-12-13 meaning if you have a right triangle with two of these sides then the unknown side is the number left out. In this case, the altitude forms a right angle on the base which allows the sides of the big isocs. triangle to be 13 giving us 13 + 13 + 10 to be the perimeter = 36.

6 0

3 years ago