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3 years ago

6

What's the midpoint between 3 and -5

1 answer:

IgorLugansk [536]3 years ago

5 0

Answer:

Step-by-step explanation:

just imagine you are on a number line and the numbers are

-5....-4.....-3.....-2....-1....0....1....2.....3

as you see the midpoint is -1

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Please help quickly

Crank

Answer:25 m

Step-by-step explanation:

3 0

3 years ago

If (x + 2) is a factor of x3 − 6x2 + kx + 10, k =

ch4aika [34]

Answer:

The value of k is -11

Step-by-step explanation:

If (x+2) is a factor of x3 − 6x2 + kx + 10:

Then,

f(x)=x3 − 6x2 + kx + 10

f(-2)=0

f(-2)=(-2)³-6(-2)²+k(-2)+10=0

f(-2)= -8-6(4)-2k+10=0

f(-2)= -8-24-2k+10=0

Solve the like terms:

f(-2)=-2k-22=0

f(-2)=-2k=0+22

-2k=22

k=22/-2

k=-11

Hence the value of k is -11....

5 0

3 years ago

How is 340 divided by 17 reasonable

vivado [14]

17 goes into 34 two times

and goes into 0 zero times

So 340 / 17 = 20

8 0

3 years ago

Read 2 more answers

10 points <br> HELP ASAP<br> PLEASE AND THANK YOU!!

BARSIC [14]

x=3

multiply 2 to both sides to cancel denominator

subtract 8x from both sides and then divide by 2

8 0

3 years ago

Read 2 more answers

Problem1 The behavior of a physical system can be described by the following first order differential equation: dy/dt=2y +t^2

daser333 [38]

Answer:

$y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}$ .

Step-by-step explanation:

Using first order linear differential equation:

$\dfrac{\mathrm{d} y}{\mathrm{d} t} = 2y + t^2$

$\frac{\mathrm{d} y}{\mathrm{d} t} - 2y =t^2$

finding integrating factor:

I.F = $e^{\int -2dt}$

I.F = $e^{-2t}$

now,

$y = \dfrac{1}{IF}(\int t^2dt+ c )$

$y = \dfrac{1}{e^{-2t}}(\int t^2dt+ c )$

$y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}$

hence the solution is

$y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}$

8 0

4 years ago