All categories

3 years ago

8% of what number is 2

Mathematics

1 answer:

Makovka662 [10]3 years ago

8 0

8% = $\frac{80}{100} = \frac{80\div4}{100\div4} = \frac{2}{25}$

$\frac{2}{25}*x=2$

$2x=25*2$
$2x=50$
$\boxed {x=25}$

You might be interested in

What is the volume, in cubic inches, of one cube with an edge length of inch?

Alex_Xolod [135]

Answer:

We know that each cube with a -inch edge length has a volume of cubic inch, because . Since the prism is built using 24 of these cubes, its volume, in cubic inches, would then be , or 3 cubic inches.

Step-by-step explanation:

7 0

2 years ago

Higher Order Thinking Some of the largest insects in the world include the Rhinoceros Beetle, the Giant Walking Stick, and the G

aleksandr82 [10.1K]

The ________ cavity contains the esophagus, trachea, bronchi, lungs, heart, and large blood vessels.
2 points
Cranial
Abdominal
Thoracic
Buccal

4 0

2 years ago

A tissue box in the shape of a rectangular prism has a surface area of 158 square inches . The width is 5 inches and the height

Agata [3.3K]

Answer:

A. 8.0 inches

you will have to use the formula A= 2(LW + LH + WL)

3 0

2 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

2 years ago

Mhanifa please help look at the picture attached

olga55 [171]

Answer:

B. (3, 2)

Step-by-step explanation:

Given vertices of triangle:

A(1, 2), B(3, 4), C(5, 0)

The centroid is found as the average of x- and y- coordinates of three vertices:

C = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)

Substitute the coordinates into formula:

C = ((1 + 3 + 5)/3, (2 + 4 + 0)/3) = (3, 2)

Correct choice is B

3 0

2 years ago

Read 2 more answers