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sleet_krkn [62]

2 years ago

11

39,892 to the nearest 1000 on the number line

1 answer:

Iteru [2.4K]2 years ago

7 0

I'm pretty sure it would be 40,000 because if i'm not mistaken you meant the nearest 10,000 since the number is 39,892.

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Free to whoever gets it

liraira [26]

LolAnswer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Round 32.697 to the nearest tenth hundredth and whole number

zalisa [80]

To the nearest tenth : 32.7

To the nearest hundredth : 32.70

Whole Number : 33

Hope this helps!

5 0

3 years ago

Need help only good answer plis!

Nesterboy [21]

A. The angles at the intersection of the two lines can be proven to be congruent and complementary . so they meet at a right angle and the lines are perpendicular.

<u>Step-by-step explanation:</u>

In above question, In order to find whether AB ⊥ CD, Using compass construction & rounder , keep the tip at A and cut arcs at line CD . Follow the same process again with tip at B and cut arcs at line CD . Do this both sides of Line CD i.e. on left side of AB & on right side of AB. Now, join the intersection points of both side arcs which are intersecting each other. Now, to prove both are right angle to each other i.e. AB ⊥ CD , can be done by proving congruent and complementary , so they meet at a right angle and Hence , the lines are perpendicular i.e. AB is inclined to CD at angle of 90°.

8 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

19, 36, 23, 42, 21, 33, 23, 24, 36, 42, 23, 37, 37, 33, 17

LekaFEV [45]

What is this suppose to mean

3 0

2 years ago