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3 years ago

Simplify -121 O-111 O 11 O-11 O 11

Mathematics

2 answers:

kramer3 years ago

4 0

simplification :

$\sqrt{ - 121}$

$\sqrt{ {i}^{2} \times 11 \times 11}$

$11i$

therefore, correct option is B

ella [17]3 years ago

4 0

The answer is B. 11i. You move 11 to the left of i because of I · 11 and so you switch it.

°.✩┈┈┈┈∘*┈┈୨♡୧┈┈*∘┈┈┈┈✩.°

Hope this helps .+(´^ω^`)+.

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Which of the following is equivalent to (5) Superscript seven-thirds?

LiRa [457]

Answer:

The correct answer is D

Step-by-step explanation:

The reason we get this answer is because when you are converting from exponential form, to radical form you always place the numerator as our constant's exponent in the radical ( is called the radicand because it is located in the radical) and the denominator in front of the radical, where it would be called the index.

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4 years ago

Read 2 more answers

1 1/2+2/5 how do I solve

raketka [301]

You can change the denominator to 10 which will get you to 55/10 + 4/10. You can add these two together to get 59/10

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3 years ago

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How to find the scale factor of two triangles

BabaBlast [244]

Answer:

Here's how to do it.

Step-by-step explanation:

The two triangles must be similar, that is, corresponding angles must be equal, as in the diagram below.

1. Find a corresponding pair of sides

Assume that ∆ABC is the original and ∆DEF is its scaled image.

Then, AB and DE are corresponding sides, with lengths of 8 and 12 units , respectively.

2. Set up a ratio

Ratio = side length in image/side length in original = 12/8

3. Simplify the ratio

The simplified ratio is the scale factor (SF).

SF = 12/8 = 3/2 = 1.5

The scale factor is 1.5.

6 0

3 years ago

Determine whether the integral converges.

Kryger [21]

You have one mistake which occurs when you integrate $\dfrac1{1-p^2}$ . The antiderivative of this is not in terms of $\tan^{-1}p$ . Instead, letting $p=\sin r$ (or $\cos r$ , if you want to bother with more signs) gives $\mathrm dp=\cos r\,\mathrm dr$ , making the indefinite integral equality

$\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C$

and then compute the definite integral from there.

$-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|$
$\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|$
$\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|$
$\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4$

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting $x+1=2\sec y$ . Then $\mathrm dx=2\sec y\tan y\,\mathrm dy$ , and the integral becomes

$\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy$
$\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy$
$\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}$
$\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}$
$\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)$
$=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2$
$=\dfrac{\ln5}4$

Another way to do this is to notice that the integrand's denominator can be factorized.

$x^2+2x-3=(x+3)(x-1)$

So,

$\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)$

There are no discontinuities to worry about since you're integrate over $[2,\infty)$ , so you can proceed with integrating straightaway.

$\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx$
$=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}$
$=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)$
$=\displaystyle\frac14\left(\ln1-\ln\frac15\right)$
$=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4$

Just goes to show there's often more than one way to skin a cat...

7 0

3 years ago

Write an equation of a line that passes through the point (-4, 5) and have a slope of 1/2

Anna [14]

Answer:

y = 1/2x + 7

Step-by-step explanation:

y = mx+ b

y = 5

x = -4

5 = 1/2(-4) + b

5 = -2 + b

7 = b

5 0

3 years ago