Someone help pleasee

Kate runs 2/3 of a mile on Mondays, Wednesday's and Thursdays. She runs 4/5 of a mile on Tuesday and Fridays. How many miles doe

padilas [110]

Answer:

Therefore, we conclute that Kate run 1 mile and 7/15 of a mile for the week.

Step-by-step explanation:

We know that Kate runs 2/3 of a mile on Mondays, Wednesdays and Thursdays. She runs 4/5 of a mile on Tuesday and Fridays. We calculate how many miles does Kate run during the week.

Therefore, we get

$\frac{2}{3}+\frac{4}{5}=\frac{10+12}{15}=\frac{22}{15}=1+\frac{7}{15}$

Therefore, we conclute that Kate run 1 mile and 7/15 of a mile for the week.

8 0

2 years ago

If A = {2, 3, 4, 5} and B = {5, 6, 7, 8}, what is A U B? {2, 3, 4, 5, 6, 7, 8) {5}

Ulleksa [173]

Hey!!!!

Here's your answer:

Solution:

A={2,3,4,5}

B={5,6,7,8}

 A U B={2,3,4,5} U {5,6,7,8}

 ={2,3,4,5,6,7,8}

In case of Union,we have to list all the elements which are present in both sets.

Hope it helps...

Good luck on your assignment

4 0

3 years ago

Which of the following expressions are equivalent to -0.5(1.7+1.7)−0.5(1.7+1.7)minus, 0, point, 5, left parenthesis, 1, point, 7

Allisa [31]

The expressions equivalent to -0.5(1.7+1.7) are -0.5(1.7) - 0.5(1.7)

and 2(-0.5(1.7))

<h3>Distributive law of expansion</h3>

Given the expression below;

-0.5 (1.7 + 1.7)

According to the distributive law;

A(B+C) = AB + AC

Expand

-0.5(1.7) + (-0.5)(1.7)

-0.5(1.7) - 0.5(1.7)

2(-0.5(1.7))

Hence the expressions equivalent to -0.5(1.7+1.7) are -0.5(1.7) - 0.5(1.7)

and 2(-0.5(1.7))

Learn more on distributive law here: brainly.com/question/25224410

#SPJ1

7 0

2 years ago

Im having trouble figuring out what this question means |7| + -4=?

OverLord2011 [107]

Answer:

3

Step-by-step explanation:

It means absolute value 7 plus -4

This means that if it were a -7, it would become positive because it is under absolute value bars...since it is positive nothing changes

7-4= 3

7 0

2 years ago

Solve the problem.

Lorico [155]

Answer:

Step-by-step explanation:

You must use a combination:

$_nC_k=\dfrac{n!}{k!(n-k)!}$

We have n = 16, k = 3.

Substitute:

$_{16}C_3=\dfrac{16!}{3!(16-3)!}=\dfrac{13!\cdot14\cdot15\cdot16}{2\cdot3\cdot13!}\qquad\text{cancel}\ 13!\\\\=\dfrac{14\cdot15\cdot16}{2\cdot3}=\dfrac{7\cdot5\cdot16}{1}=560$

5 0

3 years ago