All categories

3 years ago

The area of circle Z is 64πft². What is the value of r? r=__ ft

Mathematics

1 answer:

Masja [62]3 years ago

3 0

A=πr²
πr²= 64π
Subtract the from both sides
r²=64
Square root both sides
r=8

You might be interested in

Emma found a 4% interest rate for a $2750 loan for a 3-year term. Blake found the same 4% interest rate on a $2750 loan for a 5-

Sidana [21]

0.4×2750×3
0.4×2750×4

6 0

3 years ago

Read 2 more answers

Please help! Q and R are mutually exclusive events. If P(Q)=0.1 and P(R)=0.45, find P(Q or R).

Natasha2012 [34]

0.45 x 0.1 = 0.045 is the correct answer. Hope I helped.

7 0

3 years ago

Read 2 more answers

Divide (6.3 × 103) by (1.8 × 105) giving your answer in Scientific Notation.

Sauron [17]

Answer:

(6.3*10³) ÷ (1.8*10⁵)

6.3*10³ = 0.063*10⁵

then:

(6,3*10³) ÷ (1.8*10⁵) = (0.063*10⁵) ÷ (1.8*10⁵)

= 0.063/1.8

= 0.035

0.035 = 3.5 *10⁻² = 35*10⁻³

6 0

3 years ago

What is polynomail area

erastova [34]

Answer:

2x²-8 if it doesnt give you x= if it gives you an x= then comment it and ill work it out.

Step-by-step explanation:

it was pretty simple

5 0

3 years ago

Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying

pashok25 [27]

Answer:

a. $w(t)=-12e^{2t}$

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

$(D^2-2D-35)=0$

By factorization method we are finding the solution

$D^2-7D+5D-35=0$

$(D-7)(D+5)=0$

Substitute each factor equal to zero

D-7=0 and D+5=0

D=7 and D=-5

Therefore ,

General solution is

$y(x)=C_1e^{7t}+C_2e^{-5t}$

Let $y_1=e^{7t} \;and \;y_2=e^{-5t}$

We have to find Wronskian

$w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}$

Substitute values then we get

$w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}$

$w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}$

$w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}$

a. $w(t)=-12e^{2t}$

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

$y(0)=C_1+C_2$

$C_1+C_2=-7$ ....(equation I)

$y'(t)=7C_1e^{7t}-5C_2e^{-5t}$

$y'(0)=7C_1-5C_2$

$7C_1-5C_2=23$ ......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminate $C_1$

Then we get $C_2=-\frac{5}{2}$

Substitute the value of $C_2$ in I equation then we get

$C_1-\frac{5}{2}=-7$

$C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}$

Hence, the general solution is

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

7 0

3 years ago