Ask question

Ask question

All categories

4 years ago

14

Solve for v. –5v = 3v − 16

2 answers:

Artyom0805 [142]4 years ago

6 0

｡☆✼★ ━━━━━━━━━━━━━━ ☾

-5v = 3v - 16

- 3v

-8v = -16

/ -8

v = 2

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

｡☆✼★ ━━━━━━━━━━━━━━ ☾

Lemur [1.5K]4 years ago

5 0

Answer:

v = 2

Step-by-step explanation:

-5v = 3v - 16

- 3v

-8v = -16

/ -8

v = 2

You might be interested in

The heights of 10 year old children has a normal probability distribution with mean of 54.6 inches and standard deviation of 5.7

Fofino [41]

Answer:

a) 0.69

The probability that a randomly selected 10-year old child will be more than 51.75 inches tall

P(X>51.75 ) = 0.6915

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

<em>Given mean of the Population = 54.6 inches</em>

<em>Given standard deviation of the Population = 5.7 inches</em>

<em>Let 'X' be the random variable of normal distribution</em>

Let 'X' = 51.75 inches

$Z = \frac{x-mean}{S.D} = \frac{51.75-54.6}{5.7} = -0.5$

<u><em>Step(ii):</em></u>-

<em>The probability that a randomly selected 10-year old child will be more than 51.75 inches tall</em>

<em>P(X>51.75 ) = P(Z>-0.5)</em>

= 1 - P( Z < -0.5)

= 1 - (0.5 - A(-0.5))

= 1 -0.5 + A(-0.5)

= 0.5 + A(0.5) (∵A(-0.5)= A(0.5)

= 0.5 +0.1915

= 0.6915

<u><em>Conclusion</em></u>:-

<em>The probability that a randomly selected 10-year old child will be more than 51.75 inches tall</em>

<em>P(X>51.75 ) = 0.6915</em>

5 0

3 years ago

Nina is bouncing on a trampoline. Her fourth bounce is 6 feet high. Each bounce afterwards is two-thirds as high as the bounce b

Anvisha [2.4K]

Multiply 6 by (2/3)^2, since there are only 2 bounces that follow after her fourth. Your answer is 6 * 4/9 = 24/9 = 8/3 = 2 2/3 feet high.

6 0

4 years ago

Read 2 more answers

Find the magnitude of the net electric field at the origin (created by charges q1 and q2).

Readme [11.4K]

The direction of the net magnetic field is $167.36^{\circ}$

<h3>How to find the magnitude?</h3>

$q_1=-4 \mathrm{nC} \text { at }(0.6,0.8)$

$q_2=6 \mathrm{nC}$ at (0.6,0)

$r_1=$ Distance of $q_1$ from origin $=\sqrt{0.6^2+0.8^2}$

$r_2=$ Distance of $q_2$ from origin $=0.6$

$\mathrm{k}=$ Coulomb constant $$=9$ times $10^9 \mathrm{Nm}^2 / \mathrm{C}^2$$

Electric field is given by

$&E_1=\frac{k q_1}{r_1^2} \\$

$&\Rightarrow E_1=\frac{9 \times 10^9 \times 4 \times 10^{-9}}{\sqrt{0.6^2+0.8^2}} \\$

$&\Rightarrow E_1=36 \mathrm{~N} / \mathrm{C}$

$&\theta=\tan ^{-1} \frac{0.8}{0.6} \\$

$&\Rightarrow \theta=53.13^{\circ} \\$

$&E_1=36 \cos 53.13^{\circ} \hat{i}+36 \sin 53.13^{\circ} \hat{j} \\$

$&\Rightarrow E_1=21.6 \hat{i}+28.8 \hat{j}$

$&E_2=\frac{k q_2}{r_2^2} \\$

$&\Rightarrow E_2=\frac{9 \times 10^9 \times 6 \times 10^{-9}}{0.6^2} \\$

$&\Rightarrow E_2=150 \mathrm{~N} / \mathrm{C}$

The charge $q_2$ is on the $\mathrm{x}$ axis itself and it is pointing towards the origin (left side) so the sign will be negative

$E_2=-150 \hat{i}$

Resultant electric field

$&E=E_1+E_2 \\$

$&\Rightarrow E=21.6 \hat{i}+28.8 \hat{j}+(-150 \hat{i}) \\$

$&\Rightarrow E=-128.4 \hat{i}+28.8 \hat{j}$

Magnitude of electric field is given by

$&|E|=\sqrt{(-128.4)^2+28.8^2} \\$

$&\Rightarrow|E|=121.6 \mathrm{~N} / \mathrm{C}$

Magnitude of the net electric field is $121.6 \mathrm{~N} / \mathrm{C}$

Direction is given by

$\theta=\tan ^{-1} \frac{128.4}{28.8}=77.36^{\circ}$

From the -x axis

$(90+77.36)^{\circ}=167.36^{\circ}$

The direction of the net magnetic field is $167.36^{\circ}$

To learn more about magnetic field, refer to:

brainly.com/question/26257705

#SPJ4

6 0

2 years ago

Which is a greater number -4 or -2 ?

NikAS [45]

Answer:

-2

Step-by-step explanation:

On the number scale, -2 is higher because it is closer to zero and positive numbers than -4.

7 0

3 years ago

Evaluate the function g(t)=23t, when t = 30.

Mademuasel [1]

The best answer would be be C.20

7 0

3 years ago