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3 years ago

-20 ? I’m a little confused on the “2x” part

Mathematics

1 answer:

Brut [27]3 years ago

3 0

whatever you do to one side, you must do to the other side to keep the equation balanced.

in order to isolate the variable (x), you need to divide by 2. Since you are dividing the right side by 2, then you must divide the left side by 2.

-10÷2 = 2÷2x

-5 = 1x

-5 = x

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HELP ASAP PLZ

marishachu [46]

Answer:

There are 604,800 possible permutations.

Step-by-step explanation:

The order in which the digits are chosen is important, which means that the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

In this question:

7 digits from a set of 10(10 digits to 0-9). So

$P_{(10,7)} = \frac{10!}{(10-7)!} = 604800$

There are 604,800 possible permutations.

4 0

3 years ago

Read 2 more answers

PLS I'LL MARK BRAINLIST EXTRA POINTS 4 To go to school each day, Zane walks 24 yards due south to the bank then 45 yards east. H

kvv77 [185]

Answer:

G. 18 yards

Step-by-step explanation:

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3 years ago

A snowflake falls 6 feet in 4 seconds. At this rate, how far does a snowflake fall in 1 minute?

antiseptic1488 [7]

A snowflake falls 90 feet in 1 minute.

60 ÷ 4 = 15

6 x 15 = 90

8 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

3 years ago

WHY IS THIS GETTING DELETED?? I NEED HELP

aliina [53]

If this exact question is repeatedly deleted, it's probably because of the ambiguity of the given equation. I see two likely interpretations, for instance:

$\dfrac{(5\times5)^k}{5^{-8}} = 5^3$

$\dfrac{5\times 5^k}{5^{-8}} = 5^3$

If the first one is what you intended, then

$\dfrac{(5\times5)^k}{5^{-8}} = \dfrac{(5^2)^k}{5^{-8}} = \dfrac{5^{2k}}{5^{-8}} = 5^{2k-(-8)} = 5^{2k+8} = 5^3$

and it follows that

2k + 8 = 3 ==> 2k = -5 ==> k = -5/2

If you meant the second one, then

$\dfrac{5\times 5^k}{5^{-8}} = \dfrac{5^1\times5^k}{5^{-8}} = \dfrac{5^{k+1}}{5^{-8}} = 5^{k+1-(-8)} = 5^{k+9} = 5^3$

which would give

k + 9 = 3 ==> k = -6

And for all I know, you might have meant some other alternative... When you can, you should include a picture of your problem.

3 0

3 years ago