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skelet666 [1.2K]
3 years ago
7

I need help with these two equations and to explain how to do it

Mathematics
1 answer:
jekas [21]3 years ago
8 0

Answer:

27)  (7 + 9) · 3 / 6 = 48/6 = 8

28) (-7)² - 3(-7)(4) = 49 - (-84) = 133

Step-by-step explanation:

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9 and 12 have a common factor of 3 so 9/12 = 3/4
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Which of the following sets is closed under multiplication? {0, 1, 2, 4} {0, 1} {1, 2}
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It would be {0,1} :)
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a brother sister and mother have picked apples. the sister picked 10 more apples than the brother and her mother picked 12 apple
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I guess the sister picked 20 apples

Step-by-step explanation:

I have no explaination

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Factor this polynomial completely.<br> x^2-8x+12
Sphinxa [80]

Answer:

(x - 2) (x - 6)

Step-by-step explanation:

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7 0
4 years ago
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Simplify the expression csc(-x)/1+tan^2x)
Charra [1.4K]
Assuming ya meant \frac{csc(-x)}{1+tan^2(x)}

to slimplify, we use a variation of the pythagorean identity and a decomposition into the sin and cos


for the pythaogreaon identity
cos^2(x)+sin^2(x)=1
divide both sides by cos^2(x)
1+tan^2(x)=sec^2(x) since \frac{sin(x)}{cos(x)}=tan(x)
subsitute
\frac{csc(x)}{sec^2(x)}

recall that csc(x)=\frac{1}{cos(x)}
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore csc(-x)=\frac{1}{cos(-x)}=\frac{1}{cos(x)}=csc(x)
so we get

\frac{csc(x)}{sec^2(x)}
decompose them into \frac{1}{cos(x)} and \frac{1}{sin^2(x)} to get \frac{\frac{1}{cos(x)}}{\frac{1}{sin^2(x)}}
multiply by \frac{sin^2(x)}{sin^2(x)} to get
\frac{sin^2(x)}{cos(x)}
we can furthur simlify to get
(\frac{sin(x)}{cos(x)})(sin(x))=tan(x)sin(x)
the expression simplifies to tan(x)sin(x)
5 0
3 years ago
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