All categories

3 years ago

I rep What is the solution of 4(2y + 1) = 2(y - 13) What is the answer

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

3 0

Answer:

- 5

Step-by-step explanation:

Step 1:

4 ( 2y + 1 ) = 2 ( y - 13 )

Step 2:

8y + 4 = 2y - 26

Step 3:

6y + 4 = - 26

Step 4:

6y = - 30

Answer:

y = - 5

Hope This Helps :)

You might be interested in

Flying with the wind, a small plane flew 404 mi in 2 h. Flying against the wind, the plane could fly only 386 mi in the same amo

Alika [10]

Answer: rate of the plane is 197.5 mph

rate of wind is 4.5 mph

Step-by-step explanation:

Let x represent the rate of the plane in calm air.

Let y represent the rate of the wind.

Flying with the wind, a small plane flew 404 mi in 2 h. This means that the total speed with which the plane flew is (x + y) mph.

Distance = speed × time

Distance travelled by the plane while flying with the wind is

404 = 2(x + y)

Dividing both sides of the equation by 2, it becomes

202 = x + y- - - - - - - - - - - 1

Flying against the wind, the plane could fly only 386 mi in the same amount of time. This means that the total speed at which the plane flew is (x - y) mph.

Distance = speed × time

Distance travelled by the plane while flying against the wind is

386 = 2(x - y)

Dividing both sides of the equation by 2, it becomes

193 = x - y- - - - - - - - - - - 2

Adding equation 1 to equation 2, it becomes

395 = 2x

x = 395/2

x = 197.5

Substituting x = 197.5 into equation 1, it becomes

202 = 197.5 + y

y = 202 - 197.5

y = 4.5

3 0

3 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

3 years ago

Read 2 more answers

Alis solution is _correct_or __incorrect______

kifflom [539]

C. 9y X y
i hope this is right aliana

hope this helps,
xXharleyquinn04Xx

6 0

3 years ago

A car is going 10 miles per hour when the driver hits the breaks. The car travels 3 feet after the brakes are applied. A little

Dahasolnce [82]

Answer:

Step-by-step explanation:

10/3=20/x

20/10=2

x/3=3*2

x=6

8 0

3 years ago

What is visible by 18 and 23

Tasya [4]

The only number I got was 1. I don’t think you divide 23 unless you want the remainder

6 0

3 years ago