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hoa [83]
3 years ago
5

Kim bought 10 used books at the yard sale.How much did she pay? Did you use addiction or multiplication to solve this this probl

em
Mathematics
1 answer:
asambeis [7]3 years ago
4 0

Answer:

this problem doesn't make sense

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What is the value of y?<br> A. 55°<br> B. 850<br> C. 10°<br> D. 110°
GaryK [48]
I may not know I am sorry ma’am for your math problem is to hard
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3 years ago
Which trigonometric function would you use to solve the problem?
umka21 [38]

you would use tan or cot

3 0
3 years ago
Use synthetic division to find P (-10) for P(x)=2x^3+14x^2-58x
Savatey [412]
The polynomial remainder theorem states that the remainder upon dividing a polynomial p(x) by x-c is the same as the value of p(c), so to find p(-10) you need to find the remainder upon dividing

\dfrac{2x^3+14x^2-58x}{x+10}

You have

..... | 2 ...  14  ... -58
-10 |    ... -20  ... 60
--------------------------
..... | 2 ...  -6  ....  2

So the quotient and remainder upon dividing is

\dfrac{2x^3+14x^2-58x}{x+10}=2x-6+\dfrac2{x+10}

with a remainder of 2, which means p(-10)=2.
5 0
3 years ago
Prove, cotx -tanx= 2 cot2x
ra1l [238]

Step-by-step explanation:

<em>cotx-</em><em> </em><em>tanx</em><em>=</em><em> </em><em>2cot2x</em>

<em>L.H.S</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>cosx</em><em>/</em><em>sinx-</em><em> </em><em>sinx</em><em>/</em><em>cosx</em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>cos</em><em>²</em><em>x-</em><em> </em><em>sin</em><em>²</em><em>x</em><em>)</em><em>/</em><em> </em><em>sinxcosx</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>[</em><em> </em><em>cos</em><em>²</em><em>x-</em><em>(</em><em>1</em><em>-cos</em><em>²</em><em>x</em><em>)</em><em>]</em><em>/</em><em>sinxcosx</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em> </em><em>2cos</em><em>²</em><em>x-1</em><em>)</em><em>/</em><em>sinxcosx</em>

<em> </em><em> </em><em> </em><em>2cosx</em><em>/</em><em> </em><em>2sinxcosx</em>

<em> </em><em> </em><em> </em><em>2cot2x</em>

4 0
3 years ago
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