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Stels [109]
3 years ago
8

This question consists of 2 parts, a and b.

Mathematics
1 answer:
Svetlanka [38]3 years ago
4 0

Answer:

  • A: 700
  • B: he put the decimal point in the wrong place

Step-by-step explanation:

Part A: The problem 49/0.07 can be rewritten as 4900/7 by multiplying numerator and denominator by 100. Then it is more clear that the correct answer is (49/7)×100 = 700.

___

Part B: I've always had trouble reading Marco's mind, so I cannot tell you what mistake he made. I can tell you the decimal point in his result is in the wrong place.

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goldfiish [28.3K]

Phillip would use 31 markers because you would divide 496 by 16 to see how many markers he would place.  Your final answer would be 31 markers.  Hope this helps you!


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3 years ago
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An isosceles right triangle has a hypotenuse of length e. Select all that are true for the triangle.
____ [38]
<h3>2 Answers: </h3>
  • Choice C) Leg length = \frac{e\sqrt{2}}{2}
  • Choice E) Area = \frac{e^2}{4}

Note: if the formulas don't load properly, then you may need to refresh the page.

======================================================

Explanation:

For any 45-45-90 right triangle, the hypotenuse H is always sqrt(2) times the leg L

In terms of symbols, we can say

H = L*\sqrt{2}

We're told the hypotenuse is e, so H = e

Let's solve for L in terms of e

H = L*\sqrt{2}\\\\e = L*\sqrt{2}\\\\L*\sqrt{2} = e\\\\L = \frac{e}{\sqrt{2}}\\\\

Now let's multiply top and bottom by \sqrt{2} to rationalize the denominator

L = \frac{e}{\sqrt{2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{2}*\sqrt{2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{2*2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{4}}\\\\L = \frac{e\sqrt{2}}{2}\\\\

If the hypotenuse is e, then the length of each leg is \frac{e\sqrt{2}}{2}

This matches with choice C, which is one of the two answers. This allows us to rule out choices A and B, as those expressions are not equivalent to the expression for choice C.

----------------

The two legs of any right triangle form the base and height. The order doesn't matter. The base is always perpendicular to the height. In a right triangle, the two legs are always perpendicular.

The base and height are \frac{e\sqrt{2}}{2} each.

The area of the triangle is...

\text{Area} = \frac{1}{2}*\text{base}*\text{height}\\\\\text{Area} = \frac{1}{2}*\frac{e\sqrt{2}}{2}*\frac{e\sqrt{2}}{2}\\\\\text{Area} = \frac{e\sqrt{2}*e\sqrt{2}}{2*2*2}\\\\\text{Area} = \frac{e*e\sqrt{2*2}}{2*2*2}\\\\\text{Area} = \frac{e^2\sqrt{4}}{2*2*2}\\\\\text{Area} = \frac{e^2*2}{2*2*2}\\\\\text{Area} = \frac{e^2}{2*2}\\\\\text{Area} = \frac{e^2}{4}\\\\

This matches with choice E, which is the other answer. Choices D and F are eliminated as they are not equivalent to the expression for choice E.

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Step-by-step explanation:

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