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Vladimir79 [104]
3 years ago
8

How do you find the area of a circle that has a radius of 4

Mathematics
2 answers:
Lisa [10]3 years ago
8 0

Work is provided in the image attached.

Although you did not include units, I just used centimeters.

However, if there were no units, then just put your answer.

If you do not understand my work,

please ask me in the comments below.

romanna [79]3 years ago
6 0

Answer: Area of the circle will be

50.288

Step-by-step explanation:

Area of a circle is calculated by A=πr^2

Where r is the radius, given as 4

π is a mathematical constant of 3.143

Slot the values into the formula:

A = 3.143 x 4^2

A = 3.143 x 16

A = 50.288

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What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?
nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

f(x)=x^5-10x^3+9x; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

f'(x)=5x^4-30x^2+9

Setting derivative equal to 0, we get

5x^4-30x^2+9=0

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376   : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = x^5- 10x^3+ 9x on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.


7 0
3 years ago
I need help this is hard for me and I have a few
VLD [36.1K]

Step-by-step explanation:

Subtracting N by T

(3a2 + 2a - 5) - (2a2 + a + 6)

Subtracting like terms

a2 + a - 11

7 0
3 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia
ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account  the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

              Xa = Xb

     -6 + (3*t) =  7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

         (6.8*t) = 13.4

                  t = 13.4/6.8

                  t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

4 0
2 years ago
Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16. If 100 people are randomly sel
laiz [17]

Answer:

0.03 is  the probability that for the sample mean IQ score is greater than 103.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 16

Sample size, n = 100

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6

P( mean IQ score is greater than 103)

P(x > 103)

P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)

= 1 - P(z \leq 1.875)

Calculation the value from standard normal z table, we have,  

P(x > 103) = 1 - 0.970 =0.03= 3\%

0.03 is  the probability that for the sample mean IQ score is greater than 103.

7 0
2 years ago
Charlie Co. owes $800,000 to Tilly, Inc. from whom Charlie buys its inventory. Which account would Charlie use to report the amo
bagirrra123 [75]

Answer:

I would use chase bank, checking account. they have really good over draft protection.

7 0
2 years ago
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