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3 years ago

14

If x = 8 units, y = 3 units, and h = 9 units, find the area of the trapezoid shown above using decomposition.

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

72

Step-by-step explanation:

because you do 8x9 and its 72

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2 Points<br> What is the slope of the line described by the equation below?<br> y=-x+8

MaRussiya [10]

Answer: -1

Step-by-step explanation:

slope formula: y=mx+b

m=slope

y=(-1)x+8

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3 years ago

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Use identities to find the values of the sine and cosine functions for the following angle measure.

puteri [66]

Using the cosine double angle formula,

$\cos 2\theta=2\cos^2 \theta-1=\frac{12}{13}\\\\2\cos^{2} \theta=\frac{25}{13}\\\\\cos^{2} \theta=\frac{25}{26}\\\\\boxed{\cos \theta=\frac{5}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

Using the Pythagorean identity,

$\sin^2 \theta+\cos^2 \theta=1\\\\\sin^2 \theta+\frac{25}{26}=1\\\\sin^2 \theta=\frac{1}{26}\\\\\boxed{\sin \theta=\frac{1}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

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1 year ago

(1+cos2x)/(1-cos2x) = cot^2x

sesenic [268]

We will turn the left side into the right side.

$\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x$

Use the identity:

$\cos 2x = \cos^2 x - \sin^2 x$

$\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x$

$\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x$

Now use the identity

$\sin^2 x + \cos^2 x = 1$ solved for sin^2 x and for cos^2 x.

$\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x$

$\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$

$\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

$\cot^2 x = \cot^2 x$

8 0

3 years ago

20 PTS IF WRONG 1 STAR, IF RIGHT 5 STARS, THANKS, AND BRAINLIEST

Tema [17]

i think its D, (change my mind)

4 0

3 years ago

1. If x=3-√3,<br> prove that x²<br> +36<br> x²<br> = x²

valentina_108 [34]

Answer:

we get $x^2=12-6\sqrt{3}$

Step-by-step explanation:

We are given: $x=3-\sqrt{3}$

We need to find $x^2$

Note: Since question is not clear, I am assuming that we need to find $x^2$

Solving:

$x=3-\sqrt{3} \\Taking \ square \ on \ both \ sides\\x^2=(3-\sqrt{3})^2\\$

We know that $(a-b)^2= a^2-2ab+b^2$

Using formula and simplifying

$x^2=(3)^2-2(3)(\sqrt{3})+(\sqrt{3})^2\\x^2=9-6\sqrt{3} +3\\x^2=12-6\sqrt{3} \\$

So, we get $x^2=12-6\sqrt{3}$

3 0

3 years ago