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2 years ago

If x = y and y=z which statement is true?

Mathematics

2 answers:

Kruka [31]2 years ago

6 0

Answer:

Step-by-step explanation:

if x = y, and y = z, you can write that as x = y = z, which means that x = z

dybincka [34]2 years ago

3 0

The x=y
Is this the the test of the question?

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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

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Help me answer 99 plz

Vitek1552 [10]

One tick behind one is 3/4, 1/8 is basically half of 1/4, so find 1/4 and find 1/8, 0 is sort of obvious but then not obvious. three ticks behind 3/4. 2 2/3, count UP from 1 1/4 to find 2 2/3. I really hope I helped! :)

5 0

3 years ago

I need help. this is the problem x^2/3+10=7x^1/3

Arlecino [84]

$\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow (x^{\frac{1}{3}})^2-7x^{\frac{1}
 {3}} +10=0 
\\\;\\
We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be:
\\\;\\
t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0$

$\it \Leftrightarrow (t-2)(t-5)=0 
\\\;\\
t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8
\\\;\\
t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5 \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125$

S = {8; 125}

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3 years ago

Xto the 2nd power - 49 = 0

AnnZ [28]

Answer:

7 to the second power... because 7 times is 49 and 49-49 is 0... so 7 to the second power

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3 years ago

Please help with A,B for 10 points.

denis-greek [22]

Answer:

y= (C -Ax)/B

x= (C -By)/A

Step-by-step explanation:

Ax+ By = C

solve for y

Subtract AX from each side

Ax+ By - Ax = C-Ax

By = C -Ax

Divide by B

By /B= (C -Ax)/B

y= (C -Ax)/B

solve for x

Subtract By from each side

Ax+ By - By = C-By

Ax = C -By

Divide by A

Ax/A= (C -By)/A

x= (C -By)/A

4 0

2 years ago

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