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3 years ago

11over 5 divided by 20 over 3

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

6 0

11/5 x 3/20
Simplify to 11/1 x 3/4
33/4

maria [59]3 years ago

4 0

11/5 divided by 20/3 so you would reciprocate the second fraction then multiply the two fractions together.

When you reciprocated you get:
11/5 x 3/20

And your answer will be:

33 over 100

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The formula F= 9 5 ( C)+32 converts a temperature reading from Celsius into Fahrenheit. What does 20° C convert to in terms of d

Helga [31]

Answer:

F = 9 / 5 * 20 + 32 = 9 * 4 + 32 = 68 deg F

Step-by-step explanation:

8 0

3 years ago

To solve for y in the equation 2 x + y = 5, subtract 2 from both sides of the equation.

steposvetlana [31]

Answer:

False.

Step-by-step explanation:

Because we have to isolate y and subtracting 2 will not put y alone on one side.

We have to subtract 2x in order to solve for y.

3 0

3 years ago

Read 2 more answers

Solve these questions (ALL)

Temka [501]

Answer:

-2x+3(-5x+1)=-2x+(-15x+3)

=-2x-15x+3

=-17x+3

-5/3a+1/8-1/6a-1/2=-5/3a-1/6a+1/8-1/2

=-40/24a-4/24a+3/24-12/24

=-44/24a-9/24

=-11/6a-3/8

3(1/5x-1/7)=3/5x-3/7

Hence, ans: C

-1.5w+7.5=1.5(-w+5)

=-1.5(w-5)

Hence, ans: A

3 0

2 years ago

If h(x) = 3x −7, find h(4). Show all work for credit!

Tema [17]

H(4) means that x = 4.

Replace x with 4.

h(4) = 3(4) - 7

h(4) = 12 - 7

h(4) = 5

Hope this helps!

5 0

4 years ago

Read 2 more answers

Find the surface area of the surface given by the portion of the paraboloid z=3x2+3y2 that lies inside the cylinder x2+y2=4. (hi

natta225 [31]

Parameterize the part of the paraboloid within the cylinder - I'll call it $S$ - by

$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle$

with $0\le u\le2$ and $0\le v\le2\pi$ . The region's area is given by the surface integral

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du$

Take $w=1+36u^2$ so that $\mathrm dw=72u\,\mathrm du$ , and the integral becomes

$=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw$
$=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}$
$=\dfrac\pi{54}(145^{3/2}-1)\approx101.522$

7 0

3 years ago