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3 years ago

Can anyone answer this question please?

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

4 0

When comparing the slopes of the functions you must first make y the subject of formula.

6x - y - 2 = 0 => y = 6x - 2 => slope is 6

2y = 3/2 x + 6 => y = 3/4 x + 3 => slpoe = 3/4

5 = 2x + 4y - 3 => 4y = -2x + 5 + 3 = -2x + 8 => y = -1/2 x + 2 => slope = -1/2

The first function is steepest followed by the second function and then the third function.

The slope is not affected by the y-intercept neither is the y-intercept affected by the slope.

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How does the value of a decimal change as you move left to right

Lapatulllka [165]

If you move the decimal from left to right, then you make it either a whole number or a percentage.

4 0

3 years ago

HELLLPP ME PLEASSEEE!!!!!

antoniya [11.8K]

The similar circles P and Q can be made equal by dilation and translation

The horizontal distance between the center of circles P and Q is 11.70 units
The scale factor of dilation from circle P to Q is 2.5

<h3>The horizontal distance between their centers?</h3>

From the figure, we have the centers to be:

P = (-5,4)

Q = (6,8)

The distance is then calculated using:

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have:

d = √(6 + 5)^2 + (8 - 4)^2

Evaluate the sum

d = √137

Evaluate the root

d = 11.70

Hence, the horizontal distance between the center of circles P and Q is 11.70 units

<h3>The scale factor of dilation from circle P to Q</h3>

We have their radius to be:

P = 2

Q = 5

Divide the radius of Q by P to determine the scale factor (k)

k = Q/P

k = 5/2

k = 2.5

Hence, the scale factor of dilation from circle P to Q is 2.5

Read more about dilation at:

brainly.com/question/3457976

8 0

2 years ago

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

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