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3 years ago

Find x in this right triangle.

Mathematics

1 answer:

maks197457 [2]3 years ago

4 0

Answer:

x = 12

Solve for x:

Using Pythagorean Theorem

a^2 + b^2 = c^2

a = 9

c = 15

b = x

9^2 + x^2 = 15^2

81 + x^2 = 225

x^2 = 144 | sqrt

x = 12

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What is 0.002 one tenth of.

jeka94

Let's say is "x".

so, "x" is the whole or 10/10 whilst 0.002 is 1/10 then, let's check

$\bf \begin{array}{ccll}
amount&fraction\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
0.002&\frac{1}{10}\\\\
x&\frac{10}{10}
\end{array}\implies \cfrac{0.002}{x}=\cfrac{\frac{1}{10}}{\frac{10}{10}}\implies \cfrac{0.002}{x}=\cfrac{\frac{1}{10}}{1}
\\\\\\
\cfrac{0.002}{x}=\cfrac{1}{10}\implies \cfrac{0.002\cdot 10}{1}=x$

7 0

3 years ago

Which is the best estimate for the diagonal of a rectangle with a width of x and a length of 2x, if x is 1.33?

german

The best estimate for the diagonal of the rectangle is 3 ⇒ B

Step-by-step explanation:

In a rectangle:

The length and the width are perpendicular
The length of its diagonal = $\sqrt{(length)^{2}+(width)^{2}}$ (the diagonal is the hypotenuse of a right triangle whose legs are the length and the width of the rectangle)

∵ The width of a rectangle = x

∵ The length of the rectangle = 2x

∵ The diagonal = $\sqrt{(length)^{2}+(width)^{2}}$

- Substitute the length by 2x and the width by x

∴ The diagonal = $\sqrt{(2x)^{2}+(x)^{2}}$

∴ The diagonal = $\sqrt{4x^{2}+x^{2}}$

∴ The diagonal = $\sqrt{5x^{2}}$

∵ x = 1.33

- Substitute the value of x by 1.33 to find the diagonal

∴ The diagonal = $\sqrt{5(1.33)^{2}}$

∴ The diagonal = 2.973

∴ The diagonal ≅ 3

The best estimate for the diagonal of the rectangle is 3

Learn more:

You can learn more about the rectangles in brainly.com/question/13084321

#LearnwithBrainly

6 0

3 years ago

A basketball court is 150 feet from the net to net how many yards make up a basketball court

Snezhnost [94]

I believe you just multiply 150x3 and get 450

6 0

4 years ago

10-4/4x89-1/2+43992/69x174

True [87]

10.127781530000000000

3 0

2 years ago

Saturn is 8.867 × 10^8 miles away from the Sun. Uranus is 1.787 × 10^9 miles away from the Sun. Approximately how many times far

Over [174]

Answer:

2 times

Step-by-step explanation:

Please refer to the image attached of our solar system in which Saturn and Uranus are clearly visible.

Distance of Saturn from Sun, $D_{1} = 8.867 \times 10^{8}\ miles$

Distance of Uranus from Sun, $D_{2} = 1.787 \times 10^{9}\ miles$

We have to find, how many times the Uranus is far from Sun than Saturn is far from Sun.

Let us begin by multiplying the distance of Saturn from Sun by 2: $D_{1} \times 2 = 8.867 \times 10^{8}\ miles \times 2\\\Rightarrow 17.734 \times 10^{8}\ miles\\\Rightarrow 1.7734 \times 10^{1} \times 10^{8}\ miles\\\Rightarrow 1.7734 \times 10^{8+1}\ miles\\\Rightarrow 1.7734 \times 10^{9} $\approx$ D_{2}$

From above, we can say that $D_{2}$ is approximately twice of $D_{1}$ .

In other words, we can say that Uranus is at approximately twice distance from Sun than that of Saturn is.

3 0

3 years ago