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kirza4 [7]
3 years ago
13

The lengths of the legs of a right triangle are 8 inches and 9 inches. What is the length of the hypotenuse? Round to the neares

t inch. Enter your answer in the box.
Mathematics
1 answer:
patriot [66]3 years ago
6 0
Hyp^2 = 8^2 + 9^2
hyp^2 = 64 + 81
hyp^2 = 145
hot = 12 inch
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The sum of two-sevenths of a number and 3 is 9. Can you please write out into an equation>~
serg [7]

Answer:

2/7x +3=9

x=21

Step-by-step explanation:

Let the unknown be x.

Take 3 to the other side so: 2/7 multiplied by x= 6

Divide 6 by 2/7 to get the answer;

Therefore x=21

Sub back into equation to check answer.

4 0
3 years ago
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89,659 to the nearest thousand
zubka84 [21]
90,000 is the answer.
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The end point P of a line segment PQ(-3,6) and its mid point is (5,8) Find the coordinates of the end point Q
Georgia [21]

Answer:

  • (13, 10)

Step-by-step explanation:

Let Q have coordinates x and y

<u>Using midpoint formula we find the values of x and y</u>

  • (- 3 + x)/2 = 5 ⇒ -3 + x = 10 ⇒ x = 10 + 3 ⇒ x = 13
  • (6 + y)/2 = 8 ⇒ 6 + y = 16 ⇒ y = 16 - 6 ⇒ y = 10
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5 0
3 years ago
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Ben scores 56 out 0f 79 marks in a maths test?
zimovet [89]
<h3>Answer:  70.9%</h3>

Work Shown:

56/79 = 0.70886 approximately

Move the decimal point 2 spots to the right to go from 0.70886 to 70.886%

That rounds to 70.9%

3 0
2 years ago
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What percentage of the semi circle is shaded? <br>​
mezya [45]

The percentage of the semicircle shaded section is approximately 23,606 %.

The percentage of the area of the semicircle is equal to the ratio of the semicircle area minus the half-cross area to the semicircle area. In other words, we have the following expression:

r = \left(\frac{A_{s}-A_{h}}{A_{s}} \right)\times 100\,\%

r = \left(1-\frac{A_{h}}{A_{s}} \right)\times 100\,\% (1)

Where:

  • A_{h} - Area of the half cross, in square centimeters.
  • A_{s} - Area of the semicircle, in square centimeters.
  • r - Percentage of the shaded section of the semicircle.

And the percentage of the shaded section is:

r = \left[1-\frac{4 \cdot (2\,cm)^{2}+4\cdot \left(\frac{1}{2} \right)\cdot (2\,cm)^{2}}{0.5\cdot \pi\cdot (16\,cm^{2}+4\,cm^{2})} \right]\times 100

r \approx 23.606\,\%

The percentage of the semicircle shaded section is approximately 23,606 %.

We kindly invite to check this question on percentages: brainly.com/question/15469506

3 0
2 years ago
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