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Andre45 [30]
3 years ago
7

Can anyone help me solve this?

Mathematics
1 answer:
irina1246 [14]3 years ago
8 0
Uuhjdnxnnxnxncndndtyttygy
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How do you find the probability? give me details please and answer my question step by step.
mart [117]

Answer:

to find theoretical probability, devide the favorable outcomes by the total possible outcomes

to find experimental probability, for example on a coin, devide how many times it landed on heads by how many times you flipped it

i hope this helps :)

8 0
4 years ago
Which expression is equivalent to x² + 2x + 2?
melisa1 [442]

Answer:

(x+1-i)(x+1+i)

Step-by-step explanation:

(x+1-i)(x+1+i)\\\\=(x+1)^2 -i^2~~~~~~~~~~~~~;[a^2 -b^2 = (a+b)(a-b)]\\\\=x^2 +2x +1 -(-1)\\\\=x^2 +2x +1+1\\\\=x^2 +2x +2

4 0
2 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
An ice-skating rink hosts different events on different days of the week they keep track of ages of the people who attended the
kirill [66]

Answer:

i think it is B

Step-by-step explanation:

4 0
3 years ago
(FIRST CORRECT FOR BRAINIEST) Solve for x
Katyanochek1 [597]

Answer:

x = 12

Step-by-step explanation:

Angles in a triangle add up to equal 180 degrees

Hence, 10x - 11 + 3x - 2 + 3x + 1 = 180

( note that we just created an equation that we can use to solve for x )

we now solve for x using the equation we created

10x - 11 + 3x - 2 + 3x + 1 = 180

step 1 combine like terms

10 + 3x + 3x = 16x

- 11 - 2 + 1 = -12

we now have 180 = 16x - 12

step 2 add 12 to each side

- 12 + 12 cancels out

180 + 12 = 192

we now have 192 = 16x

step 3 divide each side by 16

16x / 16 = x

192 / 16 = 12

we're left with x = 12

4 0
3 years ago
Read 2 more answers
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