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3 years ago

AB has endpoints at A(2,6) and B(6,8). What are the coordinates of the midpoint

Mathematics

1 answer:

Trava [24]3 years ago

6 0

The answer is (4,7)

look at the attached picture

Hope it helps..

Good luck on your assignment

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A shopper uses a department-store gift card to pay for 3 sweaters priced at $27 each. What is the change in the value of the gif

LenKa [72]

Well you multiply 27 by 3 to get 71 dollars

7 0

3 years ago

A 9V battery is attached to a hand generator. What prediction best explains what will happen and why?

insens350 [35]

A hand generator it most likely will provide power to perform wok more efficiently.

<h3>A hand generator and a 9v battery</h3>

Generally, A hand generator is a portable device that use energy to do work or increase efficiency of work done.

Therefore,if a 9v battery( a battery is defined as a source of electric power) is attached to a hand generator it most likely will provide power to perform wok more efficiently

For more information on work

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3 0

3 years ago

Given g(x)=-x+3. solve for x when g(x)=-1

kherson [118]

Answer:

Step-by-step explanation:

- x + 3 = - 1 ⇒ x = 4

5 0

3 years ago

Read 2 more answers

What is the solution to 3x+30+x=10+2x+5x+2?

solmaris [256]

Answer:

answer is x = 6

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I hope it helps you

:-))

4 0

3 years ago

Read 2 more answers

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0

2 years ago