All categories

3 years ago

What's 2+2? A) 5 B) 2 C) 4 D) 10

Mathematics

1 answer:

Triss [41]3 years ago

3 0

Answer:

C) 4

Step-by-step explanation:

I hope this helps :)

You might be interested in

What is a variable? HELP HELP

bonufazy [111]

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷ A variable is an unknown number that is represented by a letter, for example, x.

It is usually found in an algebraic expression.

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

TROLLER

5 0

3 years ago

Please help I can’t figure this question out

densk [106]

I think the answer would be 7/50

8 0

2 years ago

Read 2 more answers

Show the work and answer for 5550÷10 to the third power

emmainna [20.7K]

Well I know the answer is 555 because when ever you are dividing by ten move the decimal place u move the decimal 1 place to the left or just take out the 0 at the end

8 0

3 years ago

Please help with this. i can even give brainliest!!!!

son4ous [18]

Answer: i think 10

Step-by-step explanation:

5 0

3 years ago

The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation

rodikova [14]

Answer: C) $y=sin^-1 x$

Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point $(1,\frac{\pi}{2})$ ,

If this point satisfies the function,

For the function $y=cos^{-1} x$

If x = 1

$y=cos^{-1}1=0$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=cos^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=cot^{-1}x$

If x = 1

$y=cot^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cot^{-1}x$ ,

⇒ The graph of $y=cot^{-1}x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=sin^{-1} x$

If x = 1

$y=sin^{-1}1=\frac{\pi}{2}$

Thus, $(1,\frac{\pi}{2})$ is satisfying function $y=sin^{-1} x$ ,

⇒ The graph of $y=sin^{-1} x$ is passing through the point $(1,\frac{\pi}{2})$ .

For the function $y=tan^{-1}x$

If x = 1

$y=tan^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=tan^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$ .

Hence, Option C is correct.

3 0

3 years ago