All categories

3 years ago

PLEASE HELP ASAP IT'S DUE TODAY Use the graph of the function below to answer the questions that follow.

Mathematics

1 answer:

svlad2 [7]3 years ago

6 0

Answer:

So for..

a. from: 2 to: 4

b. from: 0 to: 2 ( I believe this one, may be incorrect- sorry if so)

c. from: 6 to: 10 ( I believe also for this one, sorry if incorrect)

d. from: 5 to: 6 ( I beleive- that is what it looks like- sorry if incorrect)

e. They are staying at a constant rate, not changing, at rest...

Hope this helps!

You might be interested in

Derek found a function that approximately models the population of iguanas in a reptile garden, where x represents the number of

serious [3.7K]

Answer:

$i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ and growth rate factor is 0.075

Step-by-step explanation:

The function that models the population of iguanas in a reptile garden is given by $i(x)=12 \times (1.9)^{x}$ , where x is the number of years.

Since, $i(x)=12 \times (1.9)^{x}$

i.e. $i(x)=12 \times (1+0.9)^{x}$ .

Therefore, the monthly growth rate function becomes,

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{x \times 12}$ .

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ .

Hence, the monthly growth rate is i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ .

Also, the growth factor is given by $\frac{0.9}{12}$ = 0.075.

Thus, the growth factor to nearest thousandth place is 0.075.

4 0

3 years ago

Which diagram represents the net of a cube? Circle all that apply.

Zigmanuir [339]

Answer:

B 100%

Step-by-step explanation:

3 0

1 year ago

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0

3 years ago

Look at this I really need help with this

Anettt [7]

$y = 7500 {(.97)}^{0} \\ = 7500(1) \\ = 7500$
D

6 0

3 years ago

If you have 8 batteries cost $40 [cost per battery]

Finger [1]

Answer:

the answer is 5 because 40 divided by 8 is 5

5 0

3 years ago

Read 2 more answers