The most accurate estimation 5+4

What is the volume of the cone below?

In-s [12.5K]

Hey there! :)

Answer:

V = 3408 units³

Step-by-step explanation:

Given:

r = 12

h = 71

Volume of a cone: V = 1/3 (bh) where b = πr²

Calculate the area of the base:

A = 12²π

A = 144π units²

Plug this into the formula for the volume of a cone:

V = 1/3 (144π · 71)

V = 1/3 (10224)

V = 3408 units³

5 0

4 years ago

Read 2 more answers

I need the complete answer to this problem, please circle the final answer and what I should write

goblinko [34]

Answer:

$f(t) = 8400(1.0018)^{12t}$

2.22%

Step-by-step explanation:

$8400 is invested in an account earning 2.2% interest (APR), compounded monthly.

So, the value of the account after t years will be given by

$f(t) = 8400 (1 + \frac{2.2}{12 \times 100} )^{12t} = 8400(1.0018)^{12t}$ {Rounded to the four decimal places} (Answer)

Now, for t = 1 year, f(1) = $8586.67

So, the percentage of growth per year (APY) = $\frac{(8586.67 - 8400) \times 100}{8400} = 2.22$ % {Rounded to the nearest hundredth of percent} (Answer)

5 0

3 years ago

I need the steps for 26

Leona [35]

Answer:

K = 51

Step-by-step explanation:

Since the triangles are similar, the angles are the same measure.

<K = <F

We can find angle F since the angles of a triangle add to 180

D + G + F = 180

59+70 +F = 180

Combine like terms

129+F = 180

Subtract 129 from each side

129-129+F = 180-129

F = 51

K = F = 51

7 0

3 years ago

Read 2 more answers

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

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

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago

Can please someone help me out and explain it step by step. Thank you

Anit [1.1K]

Answer: $x=7, \ y=-2$

Step-by-step explanation:

Given

Equations are $3x-y=23$ and $2x+3y=8$

take $3x-y=23$

$y=3x-23$

Put the value of y in another equation

$\Rightarrow 2x+3(3x-23)=8\\\Rightarrow 2x+9x-69=8\\\Rightarrow 11x=77\\\Rightarrow x=7$

Put the value of x in y

$y=3(7)-23=21-23=-2$

Hence, $x=7, \ y=-2$

7 0

3 years ago