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3 years ago

Help asap!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

worty [1.4K]3 years ago

6 0

Step-by-step explanation:

${x}^{2} = 2 \times {4}^{2} \\ {x}^{2} = 32 \\ x = 4 \sqrt{2} \\ diameter = 4 \sqrt{2} \\ radius = 2 \sqrt{2} \\ area \: of \: half \: of \: the \: circle = \\$

$a = \frac{\pi \times {(2 \sqrt{2} )}^{2} }{2} = 4\pi \\ area \: of \: triangle \: = \\ 2 \sqrt{2} \times 2 \sqrt{2} = 8 \\ area \: of \: shaded \: region \: is \: = \\ 4\pi - 8 = 4.57 \: {cm}^{2}$

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A number to the 8th power divided by the same number to the 5th power equals 27. What is the number?

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x=3 when you flip the equation (rounded number)

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3 years ago

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I’m stuck on this please someone help

kakasveta [241]

Answer: $\bold{a_{n+1}=\dfrac{1}{2}a_n}$

Step-by-step explanation:

Term-to-term rule is also called the recursive rule.

The recursive rule for a geometric sequence is: $a_{n+1}=a_n\cdot r$ where

$a_n\ \text{is the previous term}$
r is the common ratio

The common ratio (r) is the second term divided by the first term (which is the same ratio for each term divided by its previous term)

In the given sequence {64, 32, 16, 8, 4}, $r=\dfrac{32}{64}\quad \rightarrow \quad r=\dfrac{1}{2}$

So, the recursive rule is: ${a_{n+1}=a_n\cdot \dfrac{1}{2}$

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3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
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So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
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where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
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3 years ago

How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com

Mnenie [13.5K]

If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)

(15 choose 2) = 105

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190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.

If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)

(15 choose 1) = 15

(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.

We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845

Therefore, there are 4845 ways to have a group of 4 with 4 women.

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee

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