What does "how many times greater" mean in math?

Solve for c: c / 3 equals 30 / 5 show your work and apply the correct order of operation

Goshia [24]

Answer:

Unless there is more to the question than c = 2 because 30 /5 = 6 and if c is 3 than 3 = 6 so simplify if needed.

3 0

3 years ago

Read 2 more answers

1. y=900(1.27)^x

Sophie [7]

Answer:

The initial value is 900

It is experiencing exponential growth by 27%

Step-by-step explanation:

Exponential functions are in the form $y=a(b)^x$ , where a is the initial value, b is the multiplier, and x is the input, such as how many years past a certain date.

Exponential growth is when the multiplier is above 1.00, or above 100%, because b is determined by 1 + r if you have exponential growth, or 1 - r if you have exponential decay. You will never use negatives with exponential decay.

7 0

2 years ago

Is (-3,4) a solution of the inequality y> - 2x – 3?

jeyben [28]

Answer:

(-3, 4) is a solution

Step-by-step explanation:

The point (-3, 4) is inside the shaded area of the graph, so is a solution.

You can check in the inequality

y > -2x -3

4 > -2(-3) -3 . . . . substitute for x and y

4 > 3 . . . . . . . true; the given point is a solution

8 0

3 years ago

Tell whether the equation has one, zero, or infinitely many solutions: 9(x - 4) +15=9x – 21

Mama L [17]

Answer:

Infinite solutions

Step-by-step explanation:

Simplify both sides, subtract 9x, and add 21x to both sides and you will get 0 = 0 so there is infinite solutions.

8 0

3 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

3 years ago