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2 years ago

Alyssa walked 1.34 fewer miles than Emily. Alyssa walked 7.45 miles. How many miles did Emily walk?

Mathematics

1 answer:

Umnica [9.8K]2 years ago

6 0

6.11 miles is Emily’s Distance amount

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Complete the paragraph to show how you can solve problems involving equivalent expressions.

miv72 [106K]

Answer:

Step-by-step explanation:

Model the expressions

with bar models or algebra tiles to determine if the expressions are

Equivalent.

You can also write equivalent expressions using

properties of operations

and order of operations.

8 0

2 years ago

Read 2 more answers

Help please I’m confused

kipiarov [429]

Answer:

h = 5.4

Step-by-step explanation:

Basically these are 2 similar triangles. That means:

Corresponding sides are proportional.

We can set up a ratio of each triangle separately and equate, cross multiply and solve for "h".

h corresponds to 2.4

2.9 corresponds to 1.3

Setting up the ratio and solving for h:

$\frac{h}{2.4}=\frac{2.9}{1.3}\\1.3h=2.4*2.9\\1.3h=6.96\\h=\frac{6.96}{1.3}\\h=5.35$

Rounded to 1 decimal place,

h = 5.4

7 0

2 years ago

(4x-3)(2x^2-7x+1) what is the product of this please help!

goblinko [34]

Answer:

$8x^{3}-34x^{2}+25x-3$

Step-by-step explanation:

In order to solve this problem, lets take the two terms in the binomial (4x and -3) and, for each one, multiply them by each term in the trinomial (2x^2, -7x, and 1):

$(4x-3)(2x^{2}-7x+1)$

$(4x*2x^{2})+(4x*-7x)+(4x*1)+(-3*2x^{2})+(-3*-7x)+(-3*1)$

$8x^{3}+(-28x^{2})+4x+(-6x^{2})+21x+(-3)\\$

$8x^{3}-28x^{2}+4x-6x^{2}+21x-3$

Let's combine like terms to get our final answer:

$8x^{3}-28x^{2}-6x^{2}+4x+21x-3$

$8x^{3}-34x^{2}+25x-3$

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B4x-1%7D%7B3%7D" id="TexFormula1" title="\frac{4x-1}{3}" alt="\frac{4x-1}{3}" align="

Anuta_ua [19.1K]

Answer:

19x-1/3

Step-by-step explanation:

it is the right answer we have to take lcm for this by taking only lcm we can solve it..

4 0

2 years ago

All four B points are equidistant from points A and C. Are they all midpoints of AC? Justify your answer.

Rina8888 [55]

Answer:

Step-by-step explanation:

For a point to be the midpoint of a line segment, it must bisect it into two equal segments and be on the line segment (hence, colinear with the endpoints). All four B points are equidistant from points A and C, but aren’t colinear with A and C. Therefore, they aren’t all midpoints of line segment AC.

I hope this helps! :)

5 0

2 years ago