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JulijaS [17]
2 years ago
8

You run 3 1/4 miles as a part of an exercise routine. You run 2/5 of that distance on a hill. You run down a slope for the last

1/2 of the hill. On how many miles of your run do you run down the slope? Write your answer as a fraction or mixed number in simplest form!
Mathematics
1 answer:
irinina [24]2 years ago
4 0

Answer:

Step-by-step explanation:

½ of ⅖ of 3¼ = ⅕ of 3¼

= ⅕ of 13/4

= 13/20

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9 – 2k = 25 what is k
worty [1.4K]

Answer:

k = - 8

Step-by-step explanation:

Subtract 9 from both sides to get the constants on one side and the variables on the other

9 - 2k = 25

- 2k = 25 - 9

Simplify

- 2k = 16

Divide by -2 on both sides to isolate the variable, then simplify once more.

- 2k/ -2 = 16/ -2

k = 16/ -2

k = -8

8 0
3 years ago
Read 2 more answers
One canned juice drink is 20% orange juice; another is 10% orange juice l. How many liters of each should be mixed together in o
Anni [7]

Answer:

There is not enough information to answer, you did not say measurements of both drinks.

Step-by-step explanation:

8 0
2 years ago
Need help asap! Show work if you can, it’s completely ok if you don’t tho
Savatey [412]
Its A) -67
plug the m and n values into the function and solve using pemdas.
5(-7)-2(-7+3)^2
-35-2(-4)^2
-35-2(16)
-35-32
-67
8 0
3 years ago
a horse weight decreases by 1/2 pound per week at this rate how many weeks will it take for the horse to visit total of 5 1/2 wr
kykrilka [37]

Answer:

11 weeks

Step-by-step explanation:

5÷1/2=10

10+1=11

Sorry if I'm wrong

4 0
2 years ago
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Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
frozen [14]

Note: Consider we need to find the vertices of the triangle A'B'C'

Given:

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).

To find:

The vertices of the triangle A'B'C'.

Solution:

If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then

(x,y)\to (y,-x)

Using this rule, we get

A(-3,6)\to A'(6,3)

B(2,9)\to B'(9,-2)

C(1,1)\to C'(1,-1)

Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).

7 0
3 years ago
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