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3 years ago

The midpoint between points (-3, 8) and (3, -4) is (0, y). What is y?

Mathematics

1 answer:

Sonbull [250]3 years ago

5 0

The midpoint between points (-3, 8) and (3, -4) is (0, 2)

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You reflect the point ( -7, -6) across

Leokris [45]

Reflecting is pushing it across from one side to another.
The answer would be (7,6)
Sorry I'm late.

4 0

3 years ago

Please help me if you can please

aleksandrvk [35]

Answer:

17. m=-1/2

18. m=-1

Step-by-step explanation:

6 0

2 years ago

Help please with this algebra problem!

natka813 [3]

F(a)=2a+8
f(a+h)=2(a+h)+8=2a+2h+8
(f(a+h)-f(a))/h=(2a+2h+8-2a-8)=2h/h=2

4 0

2 years ago

Calculate: ㅤ <img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Crightarrow%20%2B%5Cinfty%7Dx%28%5Csqrt%7Bx%5E%7B2%7D-1%7D-x%29"

RSB [31]

Answer:

$\displaystyle \large \boxed{ \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}=-\dfrac{1}{2}}$

Step-by-step explanation:

Hello, please consider the following.

$\sqrt{(x^2-1)}-x\\\\=\sqrt{x^2(1-\dfrac{1}{x^2})}-x\\\\=x\left( \sqrt{1-\frac{1}{x^2}}-1\right)$

For x close to 0, we can write

$\sqrt{1+x}=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+o(x^2)\\\\\ \text{x tends to } +\infty \text{ means }\dfrac{1}{x} \text{ tends to 0}\\\\\text{So, when }\dfrac{1}{x}\text{ is close to 0, we can write.}\\\\\sqrt{1-\dfrac{1}{x^2}}=1-\dfrac{1}{2}\dfrac{1}{x^2}-\dfrac{1}{8}\dfrac{1}{x^4}+o(\dfrac{1}{x^4})$

So,

$x\left( \sqrt{1-\frac{1}{x^2}}-1\right)\\\\=x(1-\dfrac{1}{2}\dfrac{1}{x^2}+o(\dfrac{1}{x^2})-1)\\\\=-\dfrac{1}{2x}+o(\dfrac{1}{x})$

It means that

$\displaystyle \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}\\\\=\lim_{x \rightarrow +\infty} {-\dfrac{x}{2x}}=-\dfrac{1}{2}$

Thank you

8 0

3 years ago

The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T

GREYUIT [131]

Answer:

8 minutes

Step-by-step explanation:

You can make a reasonable guess at the answer just using logic.

If the water cools at the constant rate of 5° per minute, it will take (90-60)/5 = 6 minutes for the water to cool to 60 °C. Since the rate of cooling decreases as the temperature decreases, it will take longer than 6 minutes to reach 60 °C.

The only answer choice that is longer than 6 minutes is 8 minutes.

_____

Newton's law of cooling results in an exponential function for the temperature. The initial temperature difference of 90-30 = 60 degrees decays to zero. We are told that 55/60 = 11/12 of that temperature difference remains after 1 minute, so the exponential equation can be written as ...

T(t) = 30 +60·(11/12)^t . . . . . T in °C and t in minutes

This can be solved to find the value of t for T=60.

60 = 30 +60(11/12)^t

30/60 = (11/12)^t . . . . . . subtract 30, divide by 60

log(1/2) = t·log(11/12) . . . take the log

log(1/2)/log(11/12) = t ≈ 7.96617 ≈ 8 . . . . minutes

3 0

2 years ago

The midpoint between points (-3, 8) and (3, -4) is (0, y). What is y?​

The midpoint between points (-3, 8) and (3, -4) is (0, y). What is y?