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Rainbow [258]
3 years ago
5

Your friend is on a weight loss program and has a goal of loosing 15 pounds. So far she has lost 3.5 pounds in the first two wee

ks. If she keeps up this rate of weight loss, how long can she expect it to take her to meet her weight loss goal?
Mathematics
2 answers:
Reptile [31]3 years ago
7 0

Lets do the math together.

3.5/2 to find the weight loss per week= 1.75.

We know that every week the friend will reduce her goal by 1.75.

15/1.75= 8.6.

This means that in 8.6 weeks the friend will lose 15 pounds.

I hope this Helps! :)

mojhsa [17]3 years ago
5 0

Hello there! It should take her 8.6 weeks.

Well, the question doesn't specify whether it wants how long it'll take her in weeks, months, or days, so let's go with weeks. To do this, find the unit rate by dividing the amount she lost by the number of weeks it took.

3.5 pounds ÷ 2 weeks = 1.75 pounds lost a week.

Now, divide the total she wants to loose (15 pounds) by the 1.75 pounds a week to see how many she looses in total.

15/1.75 = 8.57142... Which is a little over 8.5 weeks, so we can round up to 8.6 weeks.

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Angle Addition Crossword
Elena-2011 [213]

Answer:

Step-by-step explanation:

1). m∠AEC = m∠AEB + m∠BEC

                  = 21° + 37°

                  = 58°

2). m∠BED = m∠BEC + m∠CED

                   = 37° + 44°

                   = 81°

3). m∠IKF = m∠IKH + m∠HKG + m∠GKF

                = m∠IKH + m∠HKG + m∠IKH [Since, ∠IKH ≅ ∠GKF]

                = 2∠IKH + m∠HKG

         103° = 2∠IKH + 41°

   2(∠IKH) = 103 - 41

 m(∠IKH) = 31°

4). m∠AED = m∠AEB + m∠BEC + m∠CED

                  = 21° + 37° + 44°

                  = 102°

5). m∠JKG = 108°

    m∠JKG = m∠JKI + m∠IKH + m∠HKG

    108° = m∠JKI + 31° + 41°

    m∠JKI = 108° - 72°

    m∠JKI = 36°

6). m∠HKF = m∠GKF + m∠HKG

                  = m∠IKH + m∠HKG [Since, m∠GKF = m∠IKH]

                  = 31° + 41°

                  = 72°

7). m∠NQO = m∠MQN = 64°

8). m∠JKF = m∠JKI + m∠IKF

                 = 36° + 103°

                  = 139°

8). m∠MQO = 2(m∠NQO)

                    = 2(64)°

                    = 128°

9). m∠LQO = 156°

    m∠LQM = m∠LQO - m∠MQO

                    = 156° - 128°

                    = 28°

10. m∠NQP = m∠NQO + m∠OQP

                   = 64° + m∠LQM [Since ∠OQP ≅ ∠LQM]

                   = 64° + 28°

                   = 92°

5 0
3 years ago
1) 8(4w - 5) =<br>2) 7 (-35 - 6) -<br>can you help me solve these​
tankabanditka [31]
1) 32w -40
2) -275

Lmk if it helped you
7 0
3 years ago
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Find the range of the following equation for the domain {-4, 0, 3} f (x) = -2x^2 -3
Tamiku [17]

The range of the function is {-34, -2, -20}

<h3>How to determine the range?</h3>

The function is given as:

f(x) = -2x^2 - 2

The domain is {-4, 0, 3}

Substitute the domain values in the function.

f(-4) = -2(-4)^2 - 2 = -34

f(0) = -2(0)^2 - 2 = -2

f(3) = -2(3)^2 - 2 = -20

This means that the range of the function is {-34, -2, -20}

Read more about range at:

brainly.com/question/10197594

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7 0
2 years ago
Find the measure of ∠2 and ∠11<br><br> ∠2=________<br><br> ∠11=________
Ber [7]

Answer:

Angle 2 is 40 degrees and angle 11 is 50 degrees

Step-by-step explanation:

7 0
3 years ago
4 1/2 + 1 + 2 + 1 1/4 + 1/2 + 2/4 how did you get the answer
Anestetic [448]
Your answer is 9 3/4.
6 0
3 years ago
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