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11111nata11111 [884]
3 years ago
9

1. Which store had the highest demand both before and after the price increase?

Mathematics
2 answers:
egoroff_w [7]3 years ago
7 0

Answer:

Store 4

Step-by-step explanation:

It is seen in the photo

and the end of the chart that store 4 has the highest

Before Price Increase

and

After Price Increase

Hope this helped!

Anna71 [15]3 years ago
6 0

Answer:

The answer is Store 4

Step-by-step explanation:

As before the price increases, Store 4 has the highest price among other Stores. After price increases, Store 4 continued to has highest price among them

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Mustafa contributes 11% of his $67,200 annual salary to his 401(k) plan. What is his pretax income?
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11% of $67,200.
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Read 2 more answers
A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro
pshichka [43]

Answer:

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed \frac{dx}{dt} = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}

(h)\frac{dh}{dt}=(x)\frac{dx}{dt}

(h)\frac{dh}{dt}=rx

\frac{dh}{dt}=\frac{rx}{h}

When automobile is 30 miles past the intersection,

For x = 30

\frac{dh}{dt}=\frac{30r}{h}

Since h=\sqrt{d^{2}+(30)^{2}}

Therefore,

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

3 0
3 years ago
Jane was asked to find the factors of 8. Here is her work: 8 × 1 = 8 8 × 2 = 16 8 × 3 = 24 Factors of 8: 8, 16, 24 Which stateme
Maru [420]

Answer:

Jane found multiples of 8.

Jane should have gotten 1, 2, 4, 8 as her answer.

Step-by-step explanation:

Factors : 1,2,4,8.

1*8=8

2*4=8

4 0
3 years ago
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