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3 years ago

Blaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Mathematics

1 answer:

Rufina [12.5K]3 years ago

7 0

Answer:

this is not even math

Step-by-step explanation:

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Please help!! Timed!! Brainliest for correct, multiple choice !

Ksenya-84 [330]

Answer:

its d)x=3.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Pls help, I've been stuck on this for a week

crimeas [40]

Answer:

$\sin(C)=\pm\sqrt{1-\cos^2(C)}$

Step-by-step explanation:

Given the Pythagorean Identity:

$\sin^2(C)+\cos^2(C)=1\\$

We want to solve for sin(C).

First, we will subtract cos²(C) from both sides:

$\sin^2(C)=1-\cos^2(C)$

Next, we will take the square root of both sides. Since we are taking an even-root, we will need to add plus/minus. Hence:

$\sin(C)=\pm\sqrt{1-\cos^2(C)}$

5 0

3 years ago

If you could explain and answer I’d appreciate

Cerrena [4.2K]

Answer:

-10

Step-by-step explanation:

5 is lower than 15 so the answer must be negative

if 15 minus 5 is 10

then 5 minus 15 has to be 10

4 0

4 years ago

Read 2 more answers

I dont understand what to do and what the answer is

hodyreva [135]

Answer:

-1

Step-by-step explanation:

slope=(y2-y1)/(x2-x1)=(2-4)/(0+2)=-2/2=-1

3 0

3 years ago

Read 2 more answers

PLEASE HELP!!

erik [133]

$2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha$

$1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}$

$3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x$

8 0

3 years ago