All categories

4 years ago

(-2/3) to 2nd Power - 3/4 (2 1/3) Will give out 50+25 pts and Brainliest Answer

Mathematics

2 answers:

natta225 [31]4 years ago

8 0

Answer:

-1.30555555556?

Step-by-step explanation:

(-2/3)^2-3/4(7/3)=-1.30555555556

trasher [3.6K]4 years ago

5 0

I think the answer is -47/36

You might be interested in

The sum of 3 times a number and 6 is 8: translate the sentence into an equation

serg [7]

Answer:

3x+6=8

Step-by-step explanation:

It says 3 times a number you know that it will be 3 times the variable. Next, it says the "sum of" so you know you will be adding 6 to the multiplication of 3 and the variable to equal 8.

4 0

3 years ago

Evaluate (8+t)³-6 when t = 2.

Sophie [7]

Answer:994

Step-by-step explanation: (8+t) -6, t=2

(8+2)^3 -6

10^3-6

1000-6

994

6 0

1 year ago

X-5+2x-7 find the vale pleaser!!

frosja888 [35]

Answer:

Step-by-step explanation:

They are the same angle so they must be equal.

So, x+5 = 2x-7

5+7 = 2x-x

12 = x

3 0

3 years ago

Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you

Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

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

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

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

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

###

9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

###

12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

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

$=\dfrac{\csc x}{\sec x}$

###

13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

###

14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

$=2$

###

15.

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

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

3 0

3 years ago

The mean of five numbers is 8.

Komok [63]

Answer:

Step-by-step explanation:

5 time 8=40

40 = total

6 times 9=54

54-40=14

I hope this will help

5 0

3 years ago