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2 years ago

Help fast!!!!!!!!!!!!!

Mathematics

1 answer:

miv72 [106K]2 years ago

4 0

Answer:

Refer the photos above and have a good day

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Question 1

kramer

Answer: The answer I got was A) 16

Step-by-step explanation:

Hope this helps!!!!

8 0

3 years ago

Read 2 more answers

The day started at 48 degrees. by 3:00 p.m., the temperature has increased by 17 degrees. by nightfall, the temperature fell 17

Lemur [1.5K]

The temp. rose by 17 degrees and then fell by 17 degrees. This means that the net change in temp over the course of the day was zero.

5 0

3 years ago

What’s 433 rounded to nearest hundreds

MAXImum [283]

Answer:

400

Step-by-step explanation:

Looking at the number 433, we have to look at the number in the tens place (in this case, 3) to determine if we will round down to 400, or up to 500.

Because the number 3 is less than 5, we will round down to 400.

6 0

3 years ago

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X - 15 = 18<br><br> What is the value of x?

RSB [31]

Answer:

x=33

Step-by-step explanation

2 step equation

add 15 to 18 because there's a minus in front of the 15

15+18=33

3 0

3 years ago

Read 2 more answers

Solve the problem, calculate the line integral of f along h

Over [174]

The curve $\mathcal H$ is parameterized by

$\begin{cases}X(t)=R\cos t\\Y(t)=R\sin t\\Z(t)=Pt\end{cases}$

so in the line integral, we have

$\displaystyle\int_{\mathcal H}f(x,y,z)\,\mathrm ds=\int_{t=0}^{t=2\pi}f(X(t),Y(t),Z(t))\sqrt{\left(\frac{\mathrm dX}{\mathrm dt}\right)^2+\left(\frac{\mathrm dY}{\mathrm dt}\right)^2+\left(\frac{\mathrm dZ}{\mathrm dt}\right)^2}\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}Y(t)^2\sqrt{(-R\sin t)^2+(R\cos t)^2+P^2}\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}R^2\sin^2t\sqrt{R^2+P^2}\,\mathrm dt$
$=\displaystyle\frac{R^2\sqrt{R^2+P^2}}2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt$
$=\pi R^2\sqrt{R^2+P^2}$

You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function $f:\mathbb R^n\to\mathbb R$ , we have gradient $\nabla f:\mathbb R^n\to\mathbb R^n$ . The theorem itself then says that the line integral of $\nabla f(x,y,z)=\mathbf f(x,y,z)$ along a curve $C$ parameterized by $\mathbf r(t)$ , where $a\le t\le b$ , is given by

$\displaystyle\int_C\mathbf f(x,y,z)\,\mathrm d\mathbf r=f(\mathbf r(b))-f(\mathbf r(a))$

Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.

But this isn't the case: we're integrating $f(x,y,z)=y^2$ , a scalar function.

7 0

3 years ago