All categories

3 years ago

SOMEONE JUST PLEASE ANSWER THIS ASAP FOR BRAINLIEST!!!

Mathematics

2 answers:

vekshin13 years ago

7 0

2k^2×(k+7)×(k-5)
that ur answer

Kazeer [188]3 years ago

6 0

$2k^{2}$ (k-5)(k-7)

Hope I could help!

You might be interested in

3/8(2x-8/3x)=21 solve the equation for x

mart [117]

Answer:

$\boxed{\bold{x=-84}}$

Explanation:

Multiply Both Sides By 8:

= $\bold{8\cdot \frac{3}{8}\left(2x-\frac{8}{3}x\right)=21\cdot \:8}$

Simplify:

= $\bold{-2x=168}$

Divide Both Sides By -2:

= $\bold{\frac{-2x}{-2}=\frac{168}{-2}}$

Simplify:

= $\bold{x=-84}$

//Mordancy //.

7 0

2 years ago

Read 2 more answers

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$

4 0

2 years ago

A negative correlation between two variables x and y usually yields a negative p-value for r.

Harman [31]

A. True if im wrong sorry hope this helped tho :p

6 0

3 years ago

If you have a bag that contains 24 marbles: 4 red, 12 blue, 6 black, 2 green. what is the probability of randomly picking one ma

seropon [69]

24 total marbles

6 are black

so you have a 6/24, reduced to 1/4 probability of picking a black one.

4 0

3 years ago

Which expression is equal to 6(-3m)4?

hichkok12 [17]

Answer: -72m

Step-by-step explanation:

Simplify

7 0

2 years ago