All categories

3 years ago

Lisa, a student in an algebra class, made the following

Mathematics

1 answer:

MissTica3 years ago

4 0

False. The only requirement for two lines to be parallel is for them to have the same slope

You might be interested in

Solve for z.<br><br> 3 + 9z = –7 + 10z<br><br> z =

stiv31 [10]

The answer should be z = 10

6 0

3 years ago

Read 2 more answers

Sphere and a cylinder have the same radius and height. The volume of the cylinder is 64 meters cubed.

Mrrafil [7]

Answer:

C. 128/3 meters cubed

Step-by-step explanation:

The volume of a cylinder is denoted by: $V=\pi r^2h$ , where r is the radius and h is the height. We know it's equal to 64, so we can set that equal to V:

$V=\pi r^2h$

$64=\pi r^2h$

We know that the sphere and cylinder have the same height and radius. However, the "height" of a sphere is actually the same as its diameter, which is twice its radius. Then, we can replace h in the above equation with 2r:

$64=\pi r^2h$

$64=\pi r^2*2r=2\pi r^3$

$\pi r^3=64/2=32$

Now, the volume of a sphere is denoted by: $V=\frac{4}{3} \pi r^3$ , where r is the radius. From above, we know that $\pi r^3=32$ , so we can plug this into the equation:

$V=\frac{4}{3} \pi r^3$

$V=\frac{4}{3} *32=128/3$

Thus, the answer is C.

4 0

3 years ago

Read 2 more answers

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.