How many solutions does this equation have 10+5x/15=4

The figure shows the graphs of the cost and revenue functions for a company that manufactures and sells small radios. More than

Lyrx [107]

ANSWER

The company must produce and sell more than 700 radios to have a profit.

EXPLANATION

The revenue function is

$R(x)=40x$
and the cost function is

$C(x)=5,600+32x$

The company will break even when total revenue is equal to the total cost.

This implies that,

$40x =5,600+32x$

We group similar terms to get,

$40x - 32x = 5600$

This simplifies to,

$8x = 5600$

Divide both sides by 8 to get,

$x = \frac{5600}{8}$

$x = 700$

The company breaks even by producing and selling 700 radios.

The company must produce and sell more than 700 radios to have a profit.

5 0

2 years ago

For her 1st birthday, Ruth's grandparents invested $1000 in an 18-year certificate for her that pays 10% compounded annually. Ho

Lyrx [107]

Answer:

The certificate will be worth $5,559.92 on Ruth's 19th birthday.

Step-by-step explanation:

The compound interest formula is given by:

$A = P(1 + \frac{r}{n})^{nt}$

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that:

$P = 1000, n = 1, r = 0.1, t = 18$

So

$A = 1000(1 + \frac{0.1}{1})^{1*18}$

$A = 5559.92$

The certificate will be worth $5,559.92 on Ruth's 19th birthday.

7 0

3 years ago

A tank contains 10 liters of pure water. Saline solution with a variable concentration 5 grams of salt per liter is pumped into

algol [13]

Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min

Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

3 0

2 years ago

If f(x)=|x+2| is changed to g(x)=3f(x)+4, how is the graph of the function transformed?

frozen [14]

Answer:

download photomath it will help u with any type of math

3 0

3 years ago

SUPER EASY FOR POINTS!!!!solve: x-3=3x-2

Savatey [412]

Answer:

Polynomial equation solver

x-3=3x-2

Standard form:

−2x − 1 = 0

Factorization:

−(2x + 1) = 0

Solutions:

x = −1

2

= -0.5

3 0

2 years ago