All categories

4 years ago

What is 3/x-5=10/x+2?

Mathematics

1 answer:

Sonja [21]4 years ago

4 0

3/x - 5 = 10/x + 2

x = -1

~Hope I helped!~

You might be interested in

2/3* (1/3+3 1/4 help pls now

ale4655 [162]

Answer:

43/18

Step-by-step explanation:

2/3(1/3+ 3 1/4)

First solve parenthesis.

2/3(1/3+ 3 1/4)

find the lowest common denominator (12) to add the two fractions.

1/3= 4/12;

3 1/4= 13/4 = 39/12

4/12+ 39/12= 43/12

2/3(43/12)

Then multiply the fractions.

2/3(43/12)

Numerator: 2*43= 86

Denominator: 3*12= 36

86/36

Last simplify your fraction.

43/18 or 2 7/18

4 0

3 years ago

The ball game took in $1,666 one Saturday. The number of $12 adult tickets was 18 more than twice the number of $5 child tickets

Anni [7]

An item is regularly priced at $30. It is on sale for 20% off the regular price. How much in dollars) is discounted from the regular price?

4 0

3 years ago

In the right triangle shown, DF=EF=3DF=EF=3D, F, equals, E, F, equals, 3.

jarptica [38.1K]

Answer:

<h2>The side DE is 4.2 units long, approximately.</h2>

Step-by-step explanation:

In the given triangle, side DE is the hypothenuse. So, we can use Pythagorean's Theorem, where both legs are 3 units long.

$DE^{2}=DF^{2}+EF^{2}\\ DE^{2}=3^{2}+3^{2}\\DE^{2}=9+9\\DE=\sqrt{18} \approx 4.2$

Therefore, the side DE is 4.2 units long, approximately.

7 0

3 years ago

I need help plzz ;)

kaheart [24]

Answer:

If i am correct,

f(4) = -8

g(-3) = 39

Step-by-step explanation:

5 0

4 years ago

∫∫(x+y)dxdy ,d là miền giới hạn bởi x²+y²=1

igor_vitrenko [27]

It looks like you want to compute the double integral

$\displaystyle \iint_D (x+y) \,\mathrm dx\,\mathrm dy$

over the region D with the unit circle x ² + y ² = 1 as its boundary.

Convert to polar coordinates, in which D is given by the set

D = {(r, θ) : 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π}

and

x = r cos(θ)

y = r sin(θ)

dx dy = r dr dθ

Then the integral is

$\displaystyle \iint_D (x+y)\,\mathrm dx\,\mathrm dy = \iint_D r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \int_0^{2\pi} \int_0^1 r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \underbrace{\left( \int_0^{2\pi}(\cos(\theta)+\sin(\theta))\,\mathrm d\theta \right)}_{\int = 0} \left( \int_0^1 r^2\,\mathrm dr \right) = \boxed{0}$

3 0

3 years ago