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3 years ago

Find the volume to the nearest cubic inch of a cone whose radius are 12 inch and whose height is 15 inch ?

Mathematics

1 answer:

zubka84 [21]3 years ago

7 0

2261.95 in³

2262 in³

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Need math help please!

loris [4]

Answer:

SAS

Step-by-step explanation:

We can automatically eliminate the HL answer choice, since the given triangles aren't right triangles. The only answer choices that could make sense would be ASA or SAS, but there is no ASA choice, so the answer would be SAS.

5 0

3 years ago

Read 2 more answers

Reporting Category: Geometry Standard: 10 G.3 Recognize and solve problems involving angles formed by transversals of coplanar l

Alexxx [7]

The measure in degrees would be 15<span />

4 0

3 years ago

Find x and y on triangle<br>Also the degree is 30 and the other thing is 7sqrt3

nydimaria [60]

Answer:

y =7

x =14

Step-by-step explanation:

Since this is a right triangle we can use trig functions

tan 30 = opp /adj

tan 30 = y/ 7 sqrt(3)

7 sqrt(3) tan 30 = y

7 sqrt(3) * 1/ sqrt(3) =t

7 =y

sin 30 = opp/ hyp

sin 30 = 7/x

x sin 30 =7

x = 7/ sin 30

x = 7 / 1/2

x = 14

6 0

3 years ago

Suppose a certain population satisfies the logistic equation given by dP

Ksenya-84 [330]

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

$\frac{dP}{dt}=10P-P^2$

with P(0)=1. We need to find the population when t=3.

Using variable separable method we get

$\frac{dP}{10P-P^2}=dt$

Integrate both sides.

$\int \frac{dP}{10P-P^2}=\int dt$ .... (1)

Using partial fraction

$\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}$

$A=\frac{1}{10},B=\frac{1}{10}$

Using these values the equation (1) can be written as

$\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt$

$\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt$

On simplification we get

$\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C$

$\frac{1}{10}(\ln \frac{P}{10-P})=t+C$

We have P(0)=1

Substitute t=0 and P=1 in above equation.

$\frac{1}{10}(\ln \frac{1}{10-1})=0+C$

$\frac{1}{10}(\ln \frac{1}{9})=C$

The required equation is

$\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})$

Multiply both sides by 10.

$\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}$

$e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}$

$\frac{P}{10-P}=\frac{1}{9}e^{10t}$

Reciprocal it

$\dfrac{10-P}{P}=9e^{-10t}$

$P(t)=\dfrac{10}{1+9e^{-10t}}$

The population when t = 3 is

$P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}$

Using calculator,

$P=9.999\approx 10$

Therefore, the population when t = 3 is 10.

8 0

3 years ago

CAN SOMEONE HELP????????????????????

Triss [41]

Answer:

no, sorry

Step-by-step explanation:

6 0

3 years ago