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2 years ago

NEED ANSWER ASAP WILL GIVE BRAINLIEST

2 answers:

8 0

The answer is d. complementary angles are always angles that add up to 90 degrees. so, you can eliminate answer A and C. the question mentions a fraction which is 1/3 so u can assume the answer should have a fraction aswell. That eliminates answer B so you're left with C

ch4aika [34]2 years ago

6 0

Answer:

Step-by-step explanation:

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-5 = 7 + -4/5x I just need answer

Goryan [66]

Answer: x=15

Answer: x=15
.......

8 0

3 years ago

Read 2 more answers

In circle o, mZWBY = 72°.

likoan [24]

Answer:

Answer is B. 36 degrees

Step-by-step explanation:

From the diagram,

angle WBY = 2 × angle WXY

angle WBY = 72°

angle WXY =

$\frac{angle \: WBY}{2}$

$= \frac{72}{2}$

$= 36 \: degrees$

7 0

3 years ago

All of the following are terminating decimals except

aalyn [17]

__
4/9= 0.444 ==> never ending, repeating
1/5= 0.2 =====> terminating
19/50= 0.38 => terminating
5/8= 0.625 ==> terminating

All of the above are terminating decimals except for 4/9!!

Memes (for stupidity) :
Everyone is entitled to be stupid,
But you abuse the privilege

You find it offensive,
I find it funny...

That's why I'm happier than you!

7 0

3 years ago

Read 2 more answers

What is the area of the figure below

jek_recluse [69]

Answer:

Option A

Step-by-step explanation:

Area of the composite figure = Area of ΔABH + Area of BCGH + Area of ΔDEF

Area of ΔABH = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(6)(8-5)$

= 9 in²

Area of the trapezoid = $\frac{1}{2}(CG+BH)(\text{Distance between these lines})$

= $\frac{1}{2}(6+4)(5)$

= 25 in²

Area of ΔDEF = $\frac{1}{2}(8)(6)$

= 24 in²

Therefore, area of the given figure = 9 + 25 + 24

= 58 in²

Option A will be the correct option.

3 0

2 years ago

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$