Pls help Solve logx^343=3

Why are bars of the histogram all the same width?

The widths of the bars are equal to ensure that numbers aren't represented improperly. If a bar was wider than another that was the same height, it could be misinterpreted as being larger.

3 0

3 years ago

7x-11=5(x-2)+2x-1= ?

yan [13]

0 = 0

Simplifying
7x + -11 = 5(x + -2) + 2x + -1

Reorder the terms:
-11 + 7x = 5(x + -2) + 2x + -1

Reorder the terms:
-11 + 7x = 5(-2 + x) + 2x + -1
-11 + 7x = (-2 * 5 + x * 5) + 2x + -1
-11 + 7x = (-10 + 5x) + 2x + -1

Reorder the terms:
-11 + 7x = -10 + -1 + 5x + 2x

Combine like terms: -10 + -1 = -11
-11 + 7x = -11 + 5x + 2x

Combine like terms: 5x + 2x = 7x
-11 + 7x = -11 + 7x

Add '11' to each side of the equation.
-11 + 11 + 7x = -11 + 11 + 7x

Combine like terms: -11 + 11 = 0
0 + 7x = -11 + 11 + 7x
7x = -11 + 11 + 7x

Combine like terms: -11 + 11 = 0
7x = 0 + 7x
7x = 7x

Add '-7x' to each side of the equation.
7x + -7x = 7x + -7x

Combine like terms: 7x + -7x = 0
0 = 7x + -7x

Combine like terms: 7x + -7x = 0
0 = 0

Solving
0 = 0

3 0

3 years ago

Helpppppp!!!!!!!! Please

stiv31 [10]

Hey there!!

We know angle DEC = angle AEB

Hence,

... 4x + 12 = 5x - 10

... - x = - 22

... x = 22

4 ( 22 ) + 12

88 + 12

100° is the answer.

Hope my answer helps!

7 0

3 years ago

In the following exercises, solve

yKpoI14uk [10]

Answer:

When the volume of a gas in a container varies inversely with the pressure on the gas and a container of nitrogen has a volume of 29.5 litres with 2000 psi. So, if the tank has a 14.7 psi pressure, then the volume of the tank would be 4013.6 litres.

Step-by-step explanation:

The volume of a gas in a container varies inversely with the pressure on the gas. If a container of nitrogen has a volume of litres with psi, what is the volume if the tank has a psi pressure.

Calculate the volume to the nearest whole number.

Use the given values and calculate the proportionality constant as well as required solution.

Step 1 of 2

Write the inversely proportionality relation between the volume and pressure of a gas in container.

$v \propto \frac{1}{p}$

Where volume is represented by v and pressure is represented by p.

Remove the proportionality and place a constant say k.

$v=\frac{k}{p}$

Substitute the given values of nitrogen that v=29.5 and

p=2000 and calculate the value of k from above equation.

$\begin{aligned}&29.5=\frac{k}{2000} \\&\Rightarrow k=29.5 \times 2000 \\&\Rightarrow k=59000\end{aligned}$

Step 2 of 2

Substitute the value p=14.7 and above calculated k value in the equation

$\begin{aligned}&v=\frac{k}{p} . \\&v=\frac{59000}{14.7} \\&\Rightarrow v=4013.6\end{aligned}$

7 0

2 years ago

PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the

drek231 [11]

Answer:

1) The height of the ball from 0 to 5 seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

6 0

3 years ago