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2 years ago

What is the radius for the circle given by the equation (x - 2)^2 + y^2 = 14?

Mathematics

1 answer:

Anarel [89]2 years ago

7 0

Easily, take the square root of the number in the right side to find the radius

$\sqrt{14}$

is the answer

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Jasmine Is organizing a carnival with games Booths she needs three volunteers per booth there are 16 booth if 62 people signed u

mr_godi [17]

14 volunteers will not be able to help.

Step-by-step explanation:

According to given statement;

One booth requires 3 Volunteers.

No. of booths = 16

Given number of volunteers = 62

We will multiply total number of booths with 3 to find the number of volunteers required.

Volunteers required = No. of booths x 3

Volunteers required = $16*3\\$

Volunteers required = 48

Subtract the required volunteers from total volunteers;

$62-48=14$

14 volunteers will not be able to help.

Keywords: Subtraction, Multiplication.

Learn more about subtraction at:

#LearnwithBrainly

4 0

3 years ago

I can't figure out how to do this problem. Could somebody help me?

Verizon [17]

start by plugging in the value you already have, in this case, 5 for y.

-4x-3(5)=17

-4x-15=17

-4x=32

x=-8

however, since a is a+1, a would be 7 (making x 8)

also plz tell me if i read this wrong and it’s actually 1,5 rather than a+1 and 5 cause if i did i’m so sorry lol.

8 0

3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago

Find the values that will make the equations true.

Alex17521 [72]

The value of a=7 will make the equation true

5 0

2 years ago

Please answer this correctly

ra1l [238]

Answer:

42 13/20 km

Step-by-step explanation:

10 3/10+9 7/20+14 7/10+ 8 9/20=41+ (6+7+14+9)/20=41 + 1 13/20= 42 13/20 km

4 0

3 years ago