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Tresset [83]
2 years ago
10

R17 is greater than or equal to 545.7

Mathematics
1 answer:
zhuklara [117]2 years ago
7 0

Answer:

r17 is that a mistake or actually the number?

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A circle inside of a square. The square has side lengths of 5 meters.
Veseljchak [2.6K]

Answer:

the area is 25/4 π

Step-by-step explanation:

diameter is 5 m

area will be 25/4π

7 0
2 years ago
Read 2 more answers
Please help me I dont know the answer
jasenka [17]

Okay so the range is the largest number - the smallest number.

So 25 - 11 = 14

The Absolute Mean Deviation is going to be a bit harder to explain, just go with it.

So we first calculate the mean.

119 ÷ 7 = 17

Now we calculate the distance each value is from 17. (SEE ATTACHMENT)

What happens next is that you take those values that are the distance from 17 and you add those up and calculate the mean of that.

28 ÷ 7 = 4

The Absolute Mean Deviation is 4.

I'll figure out the bottom in a minute. But this is all I got for you so far.

5 0
3 years ago
Would it be a b c or s please answer fast
Natali [406]

Answer:

Step-by-step explanation:

it is your first choice

-4.4+3.6\geq1.6x

-0.5\geqx

7 0
3 years ago
Find the absolute maximum and minimum values of the function below. f(x) = x3 − 9x2 + 3, − 3 2 ≤ x ≤ 12 Solution Since f is cont
Neko [114]

Answer:

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

Step-by-step explanation:

The correct statement is described below:

Find the absolute maximum and minimum values of the function below:

f(x) = x^{3}-9\cdot x^{2}+ 3, 2 \leq x \leq 12

Given that function is a polynomial, then we have the guarantee that function is continuous and differentiable and we can use First and Second Derivative Tests.

First, we obtain the first derivative of the function and equalize it to zero:

f'(x) = 3\cdot x^{2}-18\cdot x

3\cdot x^{2}-18\cdot x = 0

3\cdot x \cdot (x-6) = 0 (Eq. 1)

As we can see, only a solution is a valid critical value. That is: x = 6

Second, we determine the second derivative formula and evaluate it at the only critical point:

f''(x) = 6\cdot x -18 (Eq. 2)

x = 6

f''(6) = 6\cdot (6)-18

f''(6) =18 (Absolute minimum)

Third, we evaluate the function at each extreme of the given interval and the critical point as well:

x = 2

f(2) = 2^{3}-9\cdot (2)^{2}+3

f(2) = -25

x = 6

f(6) = 6^{3}-9\cdot (6)^{2}+3

f(6) = -105

x = 12

f(12) = 12^{3}-9\cdot (12)^{2}+3

f(12) = 435

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

6 0
3 years ago
A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a
Temka [501]
K · ( 0, 0 ) = ( 0 , 0 )
k · ( 0 , 4 ) = ( 0 , 10 )
k · ( 6, 0 ) = ( 15, 0 )
k = 15/6 = 10/4 = 2.5
The answer would be: 2.5
8 0
3 years ago
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