All categories

4 years ago

Solve for x –30 = 5(x + 1)

2 answers:

6 0

-30=(5)(x)+(5)(1) (Distribute)
-30=5x+5
Flip equation
5x+5=-30
Subtract 5 from both sides
5x+5-5=-30-5
5x=-35
Divide both sides by 5
X=-7

balandron [24]4 years ago

5 0

Answer:

-30=5(x+1) is -7

Explanation:

distribute flip subtract 5 from both sides divide both sides by 5

You might be interested in

Gawaingnpang nkabuhayan,hamon at oportinidad

Savatey [412]

gawaingnpang nkabuhayan, hamon at oportinidad

7 0

3 years ago

If we increase the speed of rotation of wheel then the sound produced by cardboard touching the spokes of the wheel will become

Anarel [89]

it will become shrill

5 0

3 years ago

You want to make a ride so you do not want to exceed 1.1g’s, if the radius of the turns are 10m, then what is the maximum speed

Citrus2011 [14]

The maximum speed is 10.4 m/s

Explanation:

For a body in uniform circular motion, the centripetal acceleration is given by:

$a=\frac{v^2}{r}$

where

v is the linear speed

r is the radius of the circular path

In this problem, we have the following data:

- The maximum centripetal acceleration must be

$a=1.1 g$

where $g=9.8 m/s^2$ is the acceleration of gravity. Substituting,

$a=(1.1)(9.8)=10.8 m/s^2$

- The radius of the turn is

r = 10 m

Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:

$v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s$

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0

4 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

4 years ago

A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 17 feet. The ball is started in moti

Romashka-Z-Leto [24]

twelve because it is the answer

5 0

3 years ago

Read 2 more answers