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mihalych1998 [28]

4 years ago

8

Please help someone, I thought it was 0 but I need two numbers?

1 answer:

Strike441 [17]4 years ago

7 0

Answer:

0 & 0

Step-by-step explanation:

x² + 3 = 3

x² = 0

Root is 0 with multiplicity 2

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Eighty percent of seniors at the local high school graduate each year. If there are 750 seniors this year, how many will graduat

yulyashka [42]

What you need to find is 80% of 750

There are many ways to figure that out but the way that I did it is to divide 750 by 100. The answer (7.5). Is 1% of 750
In all I had to do was multiply 7.5 x 80. Giving us the answer which is 600

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3 years ago

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Nimfa-mama [501]

Step-by-step explanation:

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Which is the best of -3/5) (17 5/6)?

Zolol [24]

Answer:-9.

Step-by-step explanation: The actual answer is 10 7/10...

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3 years ago

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Solve these algebraic expression and make it them into one simplified expression

Scorpion4ik [409]

Step-by-step explanation:

Equation 1

$\frac{m^2}{6} -\frac{9}{4m}\\\\ by\ taking\ LCM\\\\ \frac{2m^3-27}{12m}$

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4 0

3 years ago

Simplify the first trigonometric expression by writing the simplified form in terms of the second expression (1/1-cosx)-(cos/1+c

garik1379 [7]

$(\frac{1}{1-cos(x)})-(\frac{cos(x)}{1+cos(x)})$

cscx is 1/sinx

maybe they want us to use pythagorean identity

$cos^2(x)+sin^2(x)=1$

I notice we have 1-cos(x) and 1+cos(x), if we multiply them, we get $1-cos^2(x)$

and if we look at the pythagorean identity and minus cos^2(x) from both sides, we get

$sin^2(x)=1-cos^2(x)$

since $csc(x)=\frac{1}{sin(x)}$ , $csc^2(x)=\frac{1}{sin^2(x)}=\frac{1}{1-cos^2(x)}$

(recall that (a-b)(a+b)=a²-b²)

match the denomenators of the original fraction

multiply first fraction by $\frac{1+cos(x)}{1+cos(x)}$ and the 2nd by $\frac{1-cos(x)}{1-cos(x)}$

$(\frac{1+cos(x)}{1-cos^2(x)})-(\frac{cos(x)(1-cos(x))}{1-cos^2(x)})=$

$((csc^2(x))(1+cos(x)))-((csc^2(x))(cos(x)-cos^2(x)))=$

$csc^2(x)+csc^2(x)cos(x)-(csc^2(x)cos(x)-csc^2(x)cos^2(x)=$

$csc^2(x)+csc^2(x)cos(x)-csc^2(x)cos(x)+csc^2(x)cos^2(x)=$

$csc^2(x)+csc^2(x)cos^2(x)=$

$csc^2(x)(1+cos^2(x))$ , hmm, to get ride of those cos(x)

look to the pythagorean identity again

$sin^2(x)+cos^2(x)=1$ , force one side into form 1+cos^2(x)

$1+cos^2(x)=2-sin^2(x)$ , recall that since csc(x)=1/sin(x), sin(x)=1/csc(x) and sin^2(x)=1/(csc^2(x))

$1+cos^2(x)=2-\frac{1}{csc^2(x)}$

subsituting

$csc^2(x)(1+cos^2(x))=$

$(csc^2(x))(2-\frac{1}{csc^2(x)})=$ distributing

$2csc^2(x)-\frac{csc^2(x)}{csc^2(x)}=$

$2csc^2(x)-1$ is the simplified expression

8 0

3 years ago