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3 years ago

HELP! WILL GIVE BRAINLIEST!

Mathematics

1 answer:

e-lub [12.9K]3 years ago

8 0

Answer:

I think it's "Construct arcs from point P that are greater than half the length of segment PQ"

Step-by-step explanation:

Because we need to use the arcs to create a perpendicular bisector to PQ, but we are not plotting any new points, so it can't be the last one

Hope this helps! please give brainliest!

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Write the equation for a circle that satisfies the given conditions.

Alisiya [41]

The standard form of a circle is:

$(x-a)^2+(y-b)^2=r^2$

Your center is $(-10, -6)$ and your radius is $9$

Plug them into the equation and you get

$(x-(-10))+(y-(-6)) = 9^2$

$(x+10)^2 + (y+6)^2=81$

The correct answer is D.

4 0

3 years ago

The diagram below shows the graph of the quadratic function f (x) = ax² + 8x + c. calculate the value for each of the following.

Alexxandr [17]

Answer:

see explanation

Step-by-step explanation:

The y- intercept is at (0, 6) ⇒ c = 6

The x- coordinate of the vertex is at the midpoint of the zeros, that is

m = $\frac{-3-1}{2}$ = $\frac{-4}{2}$ = - 2

Since the zeros are x = - 3 and x = - 1 then the factors are (x + 3), (x + 1) and

f(x) = a(x + 3)(x + 1) = a(x² + 4x + 3)

Comparing with f(x) = ax² + 8x + 6 ⇒ a = 2

Thus

f(x) = 2x² + 8x + 6

Substitute m = - 2 into the equation for value of n

n = 2(- 2)² + 8(- 2) + 6 = 8 - 16 + 6 = - 2

Hence

c = 6, m = - 2, a = 2, n = - 2

3 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago

If y varies directly as x, what is the value of y in these ordered pairs?<br> (8, 4) and (4, y)

soldi70 [24.7K]

So if we see the first one (8,4) x=8 and y=4

And we see the 2nd one (4,y) x=4 and y=?

We notice that the value of x its been halved. So y varies with x so y value also has to be halved so y= 2

6 0

3 years ago

A definite integral of f(x)dx on the interval a to b probably SHOULDN'T be used:

OLga [1]

= cos( -x ) * [ - sin( - x ) ] = cosx * sinx ;
cosx * sin x is the right answer !

6 0

3 years ago