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Gnoma [55]
3 years ago
13

Please help as soon as possible

Mathematics
1 answer:
ch4aika [34]3 years ago
8 0

To solve this problem you must apply the proccedure shown below:

You must switch the variables, and then, you must solve for y, as following:

x=-\frac{1}{2} \sqrt{y} +3\\ (x-3)^{2} = (\frac{1}{2}\sqrt{y})^{2}   \\ y=4x^{2} -24x+36\\ f^{-1}  (x)=4x^{2} -24x+36

Therefore, as you can see, the answer is: f^{-1} (x)= 4x^{2} -24x+36


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Can someone help me with this question??
maksim [4K]

Answer:

QT < TR.

Step-by-step explanation:

QT = TR because TZ is a mid segment so QT < TR is false.

6 0
3 years ago
An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 110 engines
slava [35]

Answer:

we will fail to reject the null hypothesis and conclude that the mean pressure is not different from 4.7 psi

Step-by-step explanation:

Let's first define the hypothesis;

Null hypothesis: H0: μ = 4.7

Alternative hypothesis: Ha: μ ≠ 4.7

We have;

Sample size; n = 110

Sample mean; x¯ = 4.6

Variance: σ² = 0.64

Standard deviation; σ = √0.64 = 0.8

Formula for test statistic is;

z = (x¯ - μ)/(σ/√n)

z = (4.6 - 4.7)/(0.8/√110)

z = -0.1/0.0763

z = -1.31

From online p-value from z-score calculator attached, using; z = -1.31, two tailed hypothesis, significance value of 0.1, we have;

P-value = 0.190196

The p-value is greater than the significance value and thus we will fail to reject the null hypothesis and conclude that the mean pressure is not different from 4.7

σ μ

8 0
2 years ago
Save me the headache
maxonik [38]

(9\sin2x+9\cos2x)^2=81

Taking the square root of both sides gives two possible cases,

9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1

or

9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1

Recall that

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

If \alpha=2x and \beta=\dfrac\pi4, we have

\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}

so in the equations above, we can write

\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1

Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi

(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

\implies x=n\pi\text{ or }\dfrac\pi4+n\pi

and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi

\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

5 0
3 years ago
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Help!!!i never learned this
andre [41]
I hope this helps you



perpendicular lines slopes multiplication -1


M1×M2= -1


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5 0
3 years ago
How many variables are in the following expression 3x+4y+z
arsen [322]

Answer:

3

Step-by-step explanation:

8 0
2 years ago
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