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dimaraw [331]
3 years ago
12

Which equation is an equation of a circle with a radius of 3 and it’s center is at (5,-2)?

Mathematics
1 answer:
storchak [24]3 years ago
5 0

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{5}{ h},\stackrel{-2}{ k})\qquad \qquad radius=\stackrel{3}{ r}\\[1em] [x-5]^2+[y-(-2)]^2=3^2\implies (x-5)^2+(y+2)^2=9

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As James bought his textbooks for classes one semester, he estimated the cost to the nearest ten dollars. He knew he could cover
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Answer:

$310

Step-by-step explanation:

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8 0
3 years ago
Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x
svetlana [45]
Hello,

sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\

= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\


==============================================================
2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\

= \dfrac{1}{sin^2x*cos^2x}} 

8 0
4 years ago
Read 2 more answers
I need help on 4 & 5 pls
Allushta [10]
What do you need help with? it helps us to know the question!
7 0
3 years ago
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