All categories

3 years ago

40% is what number is 82?

Mathematics

2 answers:

Licemer1 [7]3 years ago

8 0

32.8 because.40•82=32.8

kicyunya [14]3 years ago

6 0

Do 10% of 82 which is 8.2
then x it by 4 which would be 32.8
<u>
</u>the answer is 32.8

#swag1famalam

You might be interested in

Solve for x and explain?

zhannawk [14.2K]

Answer:

x = 7

Step-by-step explanation:

When 2 secants are drawn from an external point to a circle.

The product of the external part and the entire secant of one secant is equal to the product of the external part and the entire secant of the other secant, that is

5(5 + x) = 6(6 + 4) ← distribute and simplify both sides

25 + 5x = 6 × 10

25 + 5x = 60 ( subtract 25 from both sides )

5x = 35 ( divide both sides by 5 )

x = 7

4 0

3 years ago

What is 62500000 in standard form

8090 [49]

<span>Not included first digit (6), there are 7 digits so

62 500 000 = 6.25×10^7</span>

5 0

4 years ago

Which expression is equivalent to 4y – 1

uysha [10]

Y-.25 is your answer.

4 0

3 years ago

Which of the following is an equation of the line through (3, -1) and (−2, 14)?

enot [183]

D.
Slope is rise over run and in this case it is -15/ 5 so when it is reduced you get -3 as your slope

5 0

3 years ago

Read 2 more answers

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago