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julsineya [31]
3 years ago
13

A new toy hits the local store. Sales (in hundreds) increase at a steady rate for several months, then decrease at about the sam

e rate. This can be modeled by the function S(m)=-0.625\left|\mathrm{m}-8\right|+5S(m)=−0.625∣m−8∣+5 In what month(s) were 400 toys sold? A) 7th and 10th B) 5th C) 5th and 11th D) 4th and 12th
Mathematics
1 answer:
Montano1993 [528]3 years ago
5 0

Answer:

Option A.

Step-by-step explanation:

It is given that, sales (in hundreds) for several months can be modeled by the function:

S(m)=-0.625|m-8|+5

We need to find the month in which 400 toys were sold.

Substitute S(m)=4 in the given function.

4=-0.625|m-8|+5

4-5=-0.625|m-8|

-1=-0.625|m-8|

\dfrac{-1}{-0.625}=|m-8|

1.6=|m-8|

Now,

\pm 1.6=m-8

1.6=m-8 or -1.6=m-8

1.6+8=m or -1.6+8=m

9.6=m or 6.4=m

Approx the value to the next whole number.

m\approx 10 or m\approx 7

It means, 400 toys were sold on 7th and 10th month.

Therefore, the correct option is A.

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Answer:

Step-by-step explanation:

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We move all terms to the left:

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We add all the numbers together, and all the variables

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We get rid of parentheses

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We add all the numbers together, and all the variables

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We move all terms containing A to the left, all other terms to the right

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Step-by-step explanation:

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How do you solve #10
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LINEAR \: \: EQUATIONS \\ \\ \\ Given \: expression \: \: - \: \\ \\ \frac{7}{3} (2x + 3) + \frac{3}{4} ( \frac{x}{5} - \frac{15}{2} ) = \frac{11}{8} \\ \\ \frac{14x}{3} + \frac{21}{3} + \frac{3x}{20} - \frac{45}{8} = \frac{11}{8} \\ \\ \frac{14x}{3} + \frac{3x}{20} + \frac{21}{3} - \frac{45}{8} = \frac{11}{8} \\ \\ \frac{289x}{60} = \frac{11}{8} + \frac{45}{8} - \frac{21}{3} \\ \\ \frac{289x}{60} = \frac{56}{8} - \frac{21}{3} \\ \\ \frac{289x}{60} = 7 - 7 = 0 \\ \\ \frac{289x}{60} = 0 \\ \\ \\ \: \: \: \:|| \: \: \: x = 0 \: \: \:|| \: \: \: \: \: \: \: \: \: \: \: \: \: Ans.
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