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Natalka [10]
3 years ago
13

For each inequality, find two For each inequality, find TWO values for that make the inequality true.

Mathematics
1 answer:
Sergio [31]3 years ago
3 0

Answer: Choice D and Choice E

Explanation:

Subtract 3 from both sides

x+3 > 70

x+3-3 > 70-3

x+0 > 67

x > 67

Any number larger than 67 will make the original inequality to be true.

Unfortunately, only E) 70 fits the description.

D) 67 won't work since 67 > 67 is false.

However, I have a feeling that the original inequality might be x+3 \ge 70. If so, then 67 would work because 67\ge 67 is true.

So I'll assume that this is the inequality you meant to type out.

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If Sarah invests $5,000 into a fund that earns 6% interest compounded annually, how long will it take for her investment to grow
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Answer:

40

Step-by-step explanation:

so next year her money would be 6% more so:

5000 + (0.6 * 5000) = (1.06) * 5000

So each year she'll have : (1.06)*(last year's money)

So we can say after n years she will have :

((1.06)^n) * 5000

So:

5000((1.06)^n)=50000

n is 39.5

So around 40

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3 years ago
The manager of a store that specializes in selling tea decides to experiment with a new blend. She will mix some Earl Grey tea t
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5 0
3 years ago
If both pairs of opposite angles of a quadrilateral are supplemental then the quadrilateral is a parallelogram
Nutka1998 [239]

Answer:

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Step-by-step explanation:

8 0
3 years ago
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A=(2.21082279...,3.996803798)
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8 0
3 years ago
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The length of a rectangle is 20 units more than its width. The area of the rectangle is x4−100.
nadezda [96]

Answer:

1. x^2-10 because the area expression can be rewritten as (x^2-10)(x^2+10)which equals (x^2-10)((x^2-10)+20).

Step-by-step explanation:

Area of the rectangle =(x^4-100)

x^4-100=(x^2)^2-10^2\\$Applying difference of two squares: a^2-b^2=(a-b)(a+b)\\(x^2)^2-10^2=(x^2-10)(x^2+10)

Since the length of a rectangle is 20 units more than its width.

Width: x^2-10\\Length=x^2+10=x^2-10+20

The correct option is therefore 1.

4 0
3 years ago
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