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3 years ago

Find the volume of the pyramid (round to hundreths place) _______ft^3

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

Answer:

I got 5ft

Step-by-step explanation:

2ft+2ft+1ft=5ft

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How old is abhasra if she was at least 46 years old five years ago

adoni [48]

Answer:

She is at least 51. She could be older.

8 0

2 years ago

Read 2 more answers

The cost CC (in dollars) of making nn watches is represented by C=15n+85C=15n+85. How many watches are made when the cost is $38

dangina [55]

$\bf \stackrel{cost}{C}=15\stackrel{watches}{n}+85\qquad \boxed{C=385}\qquad \stackrel{cost}{385}=15n+85
\\\\\\
380=15n\implies \cfrac{380}{15}=n\implies \cfrac{76}{3}=n\implies 25\frac{1}{3}=n$

so, 25 whole watches and hmmmm just the wristband of another maybe.

5 0

3 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

Please help me !!! i will mark brain-list for the correct answer !!!

Shtirlitz [24]

Answer:

Step-by-step explanation:

So remember that a quarter of a circle is 90 degrees, a half of a circle is 180 degrees, and a whole circle is 360 degrees.

Looking at the image shown, it must be half a circle, whihc is 180 degrees. The image tells us that the 1st angle is 7 degrees. Now, we must find the 2nd angle.

Heres what we know however:

angle 1 + angle 2 = 180 degrees.

How do we know this?

Well, there are only 2 angles in this 180 degrees. We know that the first one is 7 degrees.

Lets input that into our nice lil equation to get an answer:

7 degrees+angle 2 = 180 degrees

Now lets solve.

Subtract 7 on both sides, and your left with:

angle 2 = 173 degrees.

So the answer must be:

<u>173 degrees</u>

Hope this helps! ;)

4 0

2 years ago

mrs.beluga is driving on a snow covered road with a drag factor of 0.2. she brakes suddenly for a deer. The tires leave a yaw ma

Aneli [31]

First, we are going to find the radius of the yaw mark. To do that we are going to use the formula: $r= \frac{c^2}{8m} + \frac{m}{2}$
where
$c$ is the length of the chord
$m$ is the middle ordinate
We know from our problem that the tires leave a yaw mark with a 52 foot chord and a middle ornate of 6 feet, so $c=52$ and $m=6$ . Lets replace those values in our formula:
$r= \frac{52^2}{8(6)} + \frac{6}{2}$
$r= \frac{2704}{48} +3$
$r= \frac{169}{3} +3$
$r= \frac{178}{3}$

Next, to find the minimum speed, we are going to use the formula: $s= \sqrt{15fr}$
where
$f$ is <span>drag factor
</span> $r$ is the radius
We know form our problem that the drag factor is 0.2, so $f=0.2$ . We also know from our previous calculation that the radius is $\frac{178}{3}$ , so $r= \frac{178}{3}$ . Lets replace those values in our formula:
$s= \sqrt{(15)(0.2)( \frac{178}{3}) }$
$s= \sqrt{178}$
$s=13.34$ mph

We can conclude that Mrs. Beluga's minimum speed before she applied the brakes was 13.34 miles per hour.

8 0

3 years ago