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2 years ago

15

If a point is selected at random in the interior of a circle, find the probability that the point is closer to the center than t

o the circle.

1 answer:

Feliz [49]2 years ago

5 0

If a point is selected randomly in the interior of a circle, the probability that the point is closer to the center than to the circle would be probably located at the area where the inner circle is or will be found.

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NO LINKS!!!!!!!!! please solve it and explain it.

Verizon [17]

Answer:

2

Explanation:

You have to calculate the rise over run. See how many times it rises from the bottom. Then see how many times it takes horizontally to get to the next corner. Not sure if this helped but I tried sorry lol (but dw the answer is correct)

3 0

3 years ago

Find the minimum value if f(x) = xe^x over [-2,0]

SashulF [63]

Given function:

$f(x)=xe^x$

The minimum value of the function can be found by setting the first derivative of the function to zero.

$f^{\prime}(x)=xe^x+e^x$ $\begin{gathered} xe^x+e^x\text{ = 0} \\ e^x(x\text{ + 1) = 0} \end{gathered}$

Solving for x:

$\begin{gathered} x\text{ + 1 = 0} \\ x\text{ = -1} \end{gathered}$ $\begin{gathered} e^x\text{ = 0} \\ \text{Does not exist} \end{gathered}$

Substituting the value of x into the original function:

$\begin{gathered} f(x=1)=-1\times e^{^{-1}}_{} \\ =\text{ -}0.368 \end{gathered}$

Hence, the minimum value in the given range is (-1, -0.368)

6 0

1 year ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago

I’m confused on this one

Lorico [155]

The question is asking which of the following does not have side that are parallel. So which of them have lines that would never touch if they were never ending lines.

4 0

3 years ago

A) Find the lowest common multiple (LCM) of 36 and 48

dezoksy [38]

144 is the lowest hoped this helped

8 0

2 years ago

Read 2 more answers