Ask question

Ask question

All categories

KonstantinChe [14]

3 years ago

8

Solve -7= sqrt 2x-9

1 answer:

Sauron [17]3 years ago

4 0

Answer:

x = 2

Explanation:

if it was -7 = the square root of both 2x-9 together, it would be false.

if it was square root of just 2x in the equation, the answer is:

x = 2

°°°°°°°°°

-7 = √2x - 9

-√2x = -9 + 7

√-2x = -2

√2x = 2

2x = 4

x = 2

You might be interested in

Unit 2 Lesson 9: Temperature & Heat Unit Test does anyone have the answers

IRISSAK [1]

Explanation:

where is the question

I did not understood this question

8 0

3 years ago

Read 2 more answers

Which sentence is correct?

Alexeev081 [22]

Answer:

pretty sure its B if it isnt im so so sorry

Explanation:

7 0

3 years ago

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$

6 0

4 years ago

How much time is needed to produce 720 Joules of work if 90 watts of power is used?

Tems11 [23]

Answer:

8 seconds

Explanation:

power (P) is defined as the rate at which work is done.

power is measured in Watts (W) , when the work done is measured in Joules (J) and time in seconds

by the definition of power,

$Power=\frac{work.done}{time.taken} \\ \\ time.taken=\frac{work.done}{power}\\=\frac{720J }{90W} \\ \\ =8 s$

3 0

3 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

Delvig [45]

Answer:

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

$E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}$

$E_w=\dfrac{\lambda }{2\pi \varepsilon _or}$

Now the electric filed due to hollow metal cylinder

$E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}$

$E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}$

Now by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

5 0

4 years ago