If r = 6 units and h = 11 units, what is the volume of the cylinder shown above? Use 3.14 for.

Does anyone know the answers to this algebra question? Will give brainiest

MaRussiya [10]

Option 1 : -22

using formula

(a + b )(a - b ) = a^2 - b^2

(√10 + 2√8) (√10 - 2√8)

= (√10)^2 - (2√8)^2

= 10 - 2×2×8=10-32

= -22

2nd question answer is option 3

3 0

2 years ago

What is an equation of the line that passes through the points (6, 0) and (-1, -7)?

Orlov [11]

Answer:

y = x - 6

Step-by-step explanation:

We want to write a line in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.

First, we need to find the slope, which is just the change in the y-coordinates divided by the change in the x-coordinates:

slope = m = $\frac{6-(-1)}{0-(-7)} =\frac{7}{7}=1$

Our equation now looks like this: y = x + b

Now, to find the y-intercept, let's use one of the points provided and plug those values of x and y into the incomplete equation we have to solve for b:

0 = 6 + b ⇒ b = -6

So, the equation is: y = x - 6.

Hope this helps!

6 0

2 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

2 years ago

from the edge of a cliff 180m high, the angles of depression of two boats at sea are 43° and 52° respectively what is the distan

MrRa [10]

tan 52 = 180/a; a = 180/tan 52

tan 43 = 180/b; b = 180/tan 43

you need to calculate, a-b = 180/tan 52 - 180/tan 43

substitute the value of tan thetas, and calculate it simply...

8 0

2 years ago

Suppose I claim that the average monthly income of all students at college is at least $2000. Express H0 and H1 using mathematic

pishuonlain [190]

Answer:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): $\mu \geq 2000$

Alternative hypothesis (H1): $\mu$

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 200 or no

Step-by-step explanation:

For this case we want to test if the the average monthly income of all students at college is at least $2000. Since the alternative hypothesis can't have an equal sign thne the correct system of hypothesis for this case are:

Null hypothesis (H0): $\mu \geq 2000$

Alternative hypothesis (H1): $\mu$

And in order to test this hypothesis we can use a one sample t or z test in order to verify if the true mean is at least 2000 or no

7 0

2 years ago