Given g(x) = x2 + 3x - 19, find g(-2)

Elena bought 320 tropical fish for a museum display. She bought 7 times as many triggerfish as parrotfish. How many of each type

luda_lava [24]

X+y = 320 y = 7x ..............

3 0

2 years ago

Which is a stretch of an exponential growth function? f(x) = Two-thirds (two-thirds) Superscript x f(x) = Three-halves (two-thir

Arte-miy333 [17]

Answer:

f(x) = Three-halves (three-halves) Superscript x

f(x) = Two-thirds (three-halves) Superscript x

Step-by-step explanation:

Since, a function in the form of $f(x) = ab^x$

Where, a and b are any constant,

is called exponential function,

There are two types of exponential function,

Growth function : If b > 1,

Decay function : if 0 < b < 1,

Since, In

$f(x) =\frac{2}{3}(\frac{2}{3})^x$

$\frac{2}{3} < 1$

Thus, it is a decay function.

in $f(x) =\frac{3}{2}(\frac{2}{3})^x$

$\frac{2}{3} < 1$

Thus, it is a decay function.

in $f(x) =\frac{3}{2}(\frac{3}{2})^x$

$\frac{3}{2} > 1$

Thus, it is a growth function.

in $f(x) =\frac{2}{3}(\frac{3}{2})^x$

$\frac{3}{2} > 1$

Thus, it is a growth function.

5 0

2 years ago

Read 2 more answers

Zane and Matt are both keen runners. Zane takes 4 minutes to jog around a running track and Matt takes 5 minutes. They start at

Sergio [31]

I believe that the correct answer to this question is B. because matt took more than 5 minutes .

3 0

2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago

Rex walks 3 1/7 km in 1 1/4 hours. At the same speed , how long will he go in one hour? Simplify your answer and write it as a p

Llana [10]

She walks 22/7 km in 5/4 hrs, so she walks (22/7)/(5/4) km, or 88/35 (2_18/35) km, in 1 hr.

7 0

2 years ago