Ask question

Ask question

All categories

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.

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

You might be interested in

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

Read 2 more answers

89,659 to the nearest thousand

zubka84 [21]

90,000 is the answer.

8 0

3 years ago

Read 2 more answers

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
Q = (13, 10)

5 0

3 years ago

Read 2 more answers

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

Read 2 more answers

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