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2 years ago

Find the value of x for which line L is parallel to M

Mathematics

1 answer:

weqwewe [10]2 years ago

4 0

Answer:

Step-by-step explanation:

Angle 95 and ( 2x - 5 ) are alternate interior angles.

Alternate interior angles are equal.

2x - 5 = 95

2x = 95 + 5

2x = 100

x = 100 / 2

x = 50

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What is the value of -2/5 divided by 6/8 PLEASE HELPPP

Tatiana [17]

Answer:

The steps to do :

$- \frac{2}{5} \div \frac{6}{8}$

$= - \frac{2}{5} \times \frac{8}{6}$

$= - \frac{1}{5} \times \frac{8}{3}$

$= - \frac{1 \times 8}{5 \times 3}$

$= - \frac{8}{15}$

7 0

3 years ago

If a number is divided by negative two, the solution is negative sixteen

Sergio [31]

N/-2=-16
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Negative number(n) is divided by negative two equals negative sixteen

3 0

3 years ago

I need help with 15 please ASAP!!!!!!

Lady bird [3.3K]

Answer:

x must equal 5 and y must equal 2

therefore,

5a + 2b = 23 and

5a - 2b = 7.

I'm gonna take 5a + 2b = 23 and add 7 to each side:

5a + 2b + 7 = 30

and substitute 5a -2b in for 7 on the left side:

5a + 2b + (5a-2b) = 30

10a = 30

a = 3

Now we know a. Now take

5a -2b = 7

and substitute in 3 for a

5*(3) - 2b = 7

15 -2b = 7

8 = 2b

b = 4

a = 3 and b = 4

Step-by-step explanation:

6 0

3 years ago

Hannah buys 3 pairs of shoes for $x each. She uses a $10 off coupon. She

iogann1982 [59]

Answer:

$15 per pair

Step-by-step explanation:

We know that the total amount of money she spent on the shoes (without discount) is 3x. Since she used a $10 off coupon, the money she spent is 3x - 10, since $10 off means - 10. And this amount equals $35:

$3x - 10 = 35\\3x = 45\\x = 45/3 \\x = 15$

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3 0

3 years ago

Read 2 more answers

A special liquid is held in a tank described as (x 2 + y 2 ) ≤ z ≤ 1 in a Cartesian coordinate system. Assume that the density o

vivado [14]

Since $\rho=\dfrac mV$ (density = mass/volume), we can get the mass/weight of the liquid by integrating the density $\rho(x,y,z)$ over the interior of the tank. This is done with the integral

$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$

which is more readily computed in cylindrical coordinates as

$\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}$

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3 years ago