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nevsk [136]
4 years ago
9

Find the distance between the two given points. Round to the nearest tenth.

Mathematics
1 answer:
labwork [276]4 years ago
5 0

Answer:

9.9

Step-by-step explanation:

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The product of 3 and a number increased by 6 is -24
padilas [110]

3(x+6)=-24 Divide both sides by 3 3(x+6)/3=-24/3 X+6=-8 X=-14

6 0
3 years ago
The radius of a semicircle is 1 kilometer. what is the semicircle's perimeter? use 3.14 for pie.
Aleksandr-060686 [28]

Answer:

5.14 km

Step-by-step explanation:

A semicircle is 1/2 of a circle so the perimeter is just 1/2 of the circumference

C = 2 * pi r for a circle so for a semicircle

1/2* 2* pi *r

pi*r

3.14 * 1

3.14

If we want to include the piece that closes the semicircle, we need to add the diameter

d = 2r = 2(1) = 2

2+3.14 = 5.14

5 0
3 years ago
Write a subtraction problem involving two positive integers with a negative difference. Explain the relationship between the two
Nadusha1986 [10]

Answer:

When both integers have the same value, the difference is zero. The difference between a positive and a negative integer can be positive or negative. When you subtract a negative integer from a positive integer, the difference is always positive this might help

4 0
3 years ago
Read 2 more answers
NEED ANSWER ASAP!!! Do the data in this table show a function? If you switch the input and the output values is it a function?
ser-zykov [4K]

Answer:

no

Step-by-step explanation:

no input can have diffrent outputs if you change the input values then yes it is a function hope this helps god bless

4 0
3 years ago
Consider the differential equation:
Wewaii [24]

(a) Take the Laplace transform of both sides:

2y''(t)+ty'(t)-2y(t)=14

\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s

where the transform of ty'(t) comes from

L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)

This yields the linear ODE,

-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s

Divides both sides by -s:

Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}

Find the integrating factor:

\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C

Multiply both sides of the ODE by e^{3\ln|s|-s^2}=s^3e^{-s^2}:

s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}

The left side condenses into the derivative of a product:

\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}

Integrate both sides and solve for Y(s):

s^3e^{-s^2}Y(s)=7e^{-s^2}+C

Y(s)=\dfrac{7+Ce^{s^2}}{s^3}

(b) Taking the inverse transform of both sides gives

y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that \frac{7t^2}2 is one solution to the original ODE.

y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7

Substitute these into the ODE to see everything checks out:

2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14

5 0
3 years ago
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