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3 years ago

Answers 1-8 ? Please and thankyou.

Mathematics

1 answer:

Licemer1 [7]3 years ago

3 0

I think the radius for #1 is 5.64 and the diameter is 11.28

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How many three-digit multiples of 5 have three different digits?

aalyn [17]

There are 180 numbers which are three-digit multiples of 5.

According to the statement

we have to find the three digit numbers which are multiples of 5.

we know that three digit numbers are start from 100 to 999 and in 1 line of counting means 100 to 110 there are 2 numbers are multiples of 5.

Here we use Multiplication method to find the number of digit which are a multiple of 5.

It means there are two numbers which are multiples of 5 in the one line of counting and there are 90 lines of counting from 100 to 999.

So, it means the answer will become

2*90 = 180.

So, There are 180 numbers which are three-digit multiples of 5.

Learn more about NUMBERS here brainly.com/question/1094036

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8 0

1 year ago

2 divided by 3460 explanation please

LUCKY_DIMON [66]

2/3460 Original Mathematical Expression.

1/1730 Divide numerator and denominator by 2.

This equals . . .

0.00057803468

(I just plugged 1/1730 into calculator)

Final Answer: 1/1730 or 0.00057803468

3 0

2 years ago

Instead of sharing with friends, Kaleb decides to share his bag of LEGO bricks with his sister, Kami. Kaleb randomly selects one

Furkat [3]

The black brick chance would be 45/117

3 0

3 years ago

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago

What is the simplified form of -(x-4) ? 0 A -X-4 B. -X+ 0 C. X+4 D. X-4

sertanlavr [38]

Hello you answer will be...... -x+4
let me know if I’m right

5 0

3 years ago