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3 years ago

Help me pleaseee!!!!

Mathematics

1 answer:

Pani-rosa [81]3 years ago

7 0

Answer: B

The diameter is 3, so you have to divide by 2 to get the raduis. After that just plug in the height and radius and that's your answer!

Hope this helped you!

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What product lies between 5x7/10=

Paraphin [41]

5/1 • 7/10=
35/10=
3 5/10=
3 1/2

5 0

3 years ago

I really need help asap show all work

Triss [41]

Note: Consider we need to find the factor form of the given expression by taking out the GCF.

Given:

The expression is:

$-42k^3+24k^4+12k^5$

To find:

The factor form of the given expression.

Solution:

We have,

$-42k^3+24k^4+12k^5$

It can be written as:

$=6k^3(-7)+6k^3(4k)+6k^3(2k^2)$

Taking out the greatest common factor (GCF), we get

$=6k^3(-7+4k+2k^2)$

Therefore, the factor form of the given expression is $6k^3(-7+4k+2k^2)$ .

8 0

3 years ago

Solve the inequality for x: 25 - 14x < 53*

Mamont248 [21]

Answer:

x must be greater than -53/14

(im not 100% sure but i think this is correct)

5 0

3 years ago

Given the line with equation 7 − 2 = 8,

timurjin [86]

Hey there!!

The basic slope-intercept formula :

y = mx+ b where ' m ' is the slope and ' b ' is the slope-intercept.

Given equation : 7x - 2y = 8

We will have to isolate the ' y '

... 7x - 2y = 8

Subtract 7x on both sides

... -2y = -7x + 8

Divide by - 2 on both sides

... y = ( 7x / 2 ) - 4

Hence,

slope = 7/2 = 3.5

y-intercept = ( 0 , -4 )

Hope it helps!!

4 0

2 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

3 years ago