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kozerog [31]
3 years ago
9

Y=3/4x+8 what is the answer

Mathematics
1 answer:
Valentin [98]3 years ago
4 0

Answer:

idk

Step-by-step explanation:

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Given that sin x = cos 2x find the value of x​
natali 33 [55]

Answer:

The most straightforward way to obtain the expression for cos(2x) is by using the "cosine of the sum" formula: cos(x + y) = cosx*cosy - sinx*siny. cos(2x) = cos² (x) - (1 - cos²(x)) = 2cos²(x) - 1. = 2cos²(x) - 1.

Step-by-step explanation:

7 0
2 years ago
What the value of 6(2b-4} when b=5
Artist 52 [7]

Answer:

6(2b - 4) \\ b = 5 \\ 6(2 \times 5 - 4) \\  = 6(10 - 4) \\ = 6(6) \\  = 36 \\ thank \: you

8 0
2 years ago
Read 2 more answers
Can y'all please help.
sattari [20]

Answer:

See explanation

Step-by-step explanation:

<u>Step 1.</u> Substitute for d to solve for n:

From the first equation:

d=70-n

Then substitute it into the second equation:

0.1(70-n)+0.05n=5.5\\ \\7-0.1n+0.05n=5.5\\ \\7+(-0.05n)=5.5\\ \\-0.05n=5.5-7=-1.5\\ \\n=\dfrac{-1.5}{-0.05}=30

<u>Step 2:</u> Substitute for n to solve for d:

d+n=70\\ \\d+30=70\\ \\d=70-30=40

There are 40 dimes and 30 nikels in the piggy bank

6 0
3 years ago
The output or possible values you could get as a result from<br> an expression or equation.
Alex73 [517]

Answer:

The Solution Set

Step-by-step explanation:

Despite missing some context, the output we could get as result of an expression/equation it is what we call a Solution set. A set which makes the initial statement of equation/expression true, therefore valid.

x^{2} -7x+12=0\\(x-3)(x-4)=0\\S=\{3,4\}

Therefore, the elements of the Solution set, 3 and 4 when plugged into the equation make the statement of the quadratic equation true.

(3)^{2}-7(3)+12=0\\-12+12=0\\0=0

4 0
3 years ago
Square roots in trigonometry. I don’t understand please help?
cupoosta [38]

By definitions of the (co)tangent and cosecant function,

3\tan^2x-2=\csc^2x-\cot^2x\iff3\dfrac{\sin^2x}{\cos^2x}-2=\dfrac1{\sin^2x}-\dfrac{\cos^2x}{\sin^2x}

Turn everything into fractions with common denominators:

\dfrac{3\sin^2x-2\cos^2x}{\cos^2x}=\dfrac{1-\cos^2x}{\sin^2x}

Recall that \cos^2x+\sin^2x=1, so we can simplify both sides a bit.

On the left:

\dfrac{3\sin^2x+3\cos^2x-5\cos^2x}{\cos^2x}=\dfrac{3-5\cos^2x}{\cos^2x}

On the right:

\dfrac{1-\cos^2x}{\sin^2x}=\dfrac{\sin^2x}{\sin^2x}=1

(as long as \sin x\neq 0, which happens in the interval 0\le x\le\pi when x=0 or x=\pi)

So we have

\dfrac{3-5\cos^2x}{\cos^2x}=1\implies3-5\cos^2x=\cos^2x

\implies3=6\cos^2x

\implies\cos^2x=\dfrac12

\implies\cos x=\pm\dfrac1{\sqrt2}

\implies x=\dfrac\pi4\text{ or }x=\dfrac{3\pi}4

4 0
3 years ago
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