Find the center and radius of a circle with the equation x² + y2 +6x

Using this distribution, find the probability

kaheart [24]

Answer:

P = 0.3

Step-by-step explanation:

Here, we are to use the probability distribution in the table to calculate the probability that a children has 4 or more shoes in his or her closet

When we say 4 or more, what we mean by this is that the teenager has 4 shoes or 5 shoes

In probability expressions, when we use the term ‘or’ we are simply talking about adding the terms involved

So what we can do here is to add the probability that the teenager has 4 shoes to the probability that the teenager has five shoes

From the table that would be; 0.1 + 0.2 = 0.3

6 0

3 years ago

How do i solve this: 5m (7-10n)=3p for n

ivanzaharov [21]

Step 1: Factor out variable m.m(−50n+35)=3pStep 2: Divide both sides by -50n+35.m(−50n+35)−50n+35=3p−50n+35m=3p−50n+35Answer:m=3p−50n+35

3 0

3 years ago

I need help, algebra hw

Bogdan [553]

The x-intercept represents the points in which the quadratic function passes through the x-axis. The maximum value represents the ordered pair with the highest range(y-value).
Interval increasing: (Negative Infinity-10)
Interval Decreasing: (10-Negative Infinity)

Part B: 8/5
Note: I am assuming the scale factor of the graph is 2.

8 0

3 years ago

How can we isolate the variable c

Vlada [557]

Well depending on the equation you look at both sides and eliminate everything from both using addition, subtraction, multiplication, and division.

Ex:

3c-4=7 first add the 4 because you use the opposite type of operation so you add it to 7 which gives you:

3c=11 next you have to divide to get it alone

3c/3=11/3

which gives you

c= 3.66666666667

8 0

3 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago

Find the center and radius of a circle with the equation x² + y2 +6x–8y +16= 0.