All categories

4 years ago

SA=πr^2-πr solve for r

1 answer:

5 0

A great time to apply the quadratic formula
for a quadratic equation
Ax^2 + Bx +C =0

then
$x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

You might be interested in

Gabriel buys 5 pounds of apples for $8. Allie buys 10 pounds of the same apples for $16. Which of the following points is NOT on

Tom [10]

I don't understand your question

4 0

4 years ago

How do you find the square root of a number?

joja [24]

By taking square root of a number using calculator

5 0

3 years ago

Operation for <br><br>w/7 = 14<br><br>addition<br>subtraction<br>multiplication<br>division

NikAS [45]

Answer: Multiplication

Step-by-step explanation:

Given

w/7=14

Multiply 7 on both sides

w/7 × 7=14 × 7

w=98

In order to isolate w on one side, we should times 7 as the numerator in order to eliminate the 7 in the denominator. To promise equality, we need to multiply 7 on both sides in order to isolate w and solve the equation.

<h2>Thus, it is multiplication.</h2>

4 0

3 years ago

Read 2 more answers

The population of a town was 9000 people (in year "0"). The population began to decrease by 3% per year. State the number of WHO

kotykmax [81]

Answer:model y = 720,500(1.022)^ , where x is the number of years ... starting value growth ratio b) What would the population be in 2000 if the growth continues at the same ... 3) A population of beetles is growing each month at a rate of 5%.

3 0

3 years ago

A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At

kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

$\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

(a) So, the probability of B,

$P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

$\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}$

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

$\implies P(A\cap B\cap C)=\frac{1}{8}$

7 0

3 years ago