The sum of the real numbers x and y is 15. Their difference is 9. What is the value of xy?

Each statement describes a transformation of the graph of f(x) = x. Which statement correctly describes the graph of g(x) if g(x

GREYUIT [131]

Answer:

D.

It is the graph of f(x) translated 3 units down.

Step-by-step explanation:

5 0

3 years ago

A plane flies from Springville to Brockton and then back to Springville. When there is no wind, the round trip takes 6 hours and

ch4aika [34]

Answer:

Step-by-step explanation:

Let x represent the speed of the plane in still air.

When there is no wind, the round trip takes 6 hours and 24 minutes. Converting to hours, it becomes

6 + 24/60 = 6 + 0.4 = 6.4 hours

Distance = speed × time

Distance of the round trip = 6.4x

But when there is a wind blowing from Springville to Brockton at 50 miles per hour, the trip takes 6 hours and 40 minutes. Converting 6 hours and 40 minutes, it becomes

6 + 40/60 = 6 2/3 = 20/3 hours

The total speed would be

(x - 50)miles per hour

Distance covered = 20/3(x - 50)

Therefore, since both round trips are the same distance, then,

6.4x = 20/3(x - 50)

Cross multiplying by 3, it becomes

19.2x = 20(x - 50)

19.2x = 20x - 1000

20x - 19.2x = 1000

0.8x = 1000

x = 1000/0.8 = 1250 mph

Distance covered on the round trip is

6.4x = 6 × 1250 = 7500 miles

Therefore, the distance from Springville to Brockton is

7500/2 = 3750 miles

7 0

3 years ago

For each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to

kiruha [24]

Answer:

D) 11/3

Step-by-step explanation:

Let x represent monthly salary.

We have been given that for each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity.

Salary earned in 11 months would be $11x$ .

Money donated in 11 months would be $\frac{11x}{10}$ .

Further we are told that in December, the worker earned N times his usual monthly salary and donated one-fifth of his earnings to charity.

Salary for December would be $Nx$ .

Money donated in December would be $\frac{Nx}{5}$ .

The worker's charitable contributions totaled one-eighth of his earnings for the entire year that is $\frac{1}{8}\cdot (11x+Nx)$ .

$\frac{11x}{10}+\frac{Nx}{5}=\frac{1}{8}\cdot (11x+Nx)$

Dividing whole equation by x, we will get:

$\frac{11x}{10x}+\frac{Nx}{5x}=\frac{1}{8x}\cdot x(11+N)$

$\frac{11}{10}+\frac{N}{5}=\frac{1}{8}\cdot (11+N)$

$\frac{11}{10}+\frac{N}{5}=\frac{11}{8}+\frac{N}{8}$

Combine like terms:

$\frac{N}{5}-\frac{N}{8}=\frac{11}{8}-\frac{11}{10}$

$\frac{N}{5}*40-\frac{N}{8}*40=\frac{11}{8}*40-\frac{11}{10}*40$

$8N-5N=11*5-11*4$

$3N=55-44$

$3N=11$

$\frac{3N}{3}=\frac{11}{3}\\\\N=\frac{11}{3}$

Therefore, the value of N is 11/3 and option D is the correct choice.

4 0

3 years ago

When you solve a problem involving money, what can a negative answer represent?

belka [17]

When solving anproblem including money a negative answer can represent debt (how much you owe)

7 0

3 years ago

Read 2 more answers

A potato was launched in the air using a potato gun. The function for the situation was h(t)=-16t^2 +100t+10, where t was the ti

Goshia [24]

Answer:

The height equation is:

h(t)=-16t^2 +100t+10

A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.

v(t) = -2*16*t + 100 = 0

t = 100/(2*16) = 3.125s

Now we put this time in our height equation:

h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.

B) The lowest heigth of the potato will be h = 0ft, when the potato hits the ground.

C) the lowest time that works for that function is t = 0s, when the potato is fired by the gun.

D) the maximum time can be finded when the potato hits the ground, afther that point the equation does not work anymore, let's find it,

h(t) = 0 = -16t^2 +100t+10

we can solve it using Bhaskara's equation:

$t = \frac{-100 +- \sqrt{100^2 +4*16*10} }{-2*16} = \frac{-100 +- 103.2}{-32}$

So the two solutions are:

t = 6.3s

t = -0.1s

We need to chose the positive time, as we already discused that the minimum time that works for the equation is t = 0s.

so here the answer is t = 6.3s

E) the domain is 0s ≤ t ≤ 6.3s

The range is 0ft ≤ h ≤ 166.25 ft.

4 0

3 years ago