$x^3+5x^2+9x+45=x^2(x+5)+9(x+5)=(x+5)(x^2+9)=(*)\\\\x^2+9=x^2+3^2=x^2-(-1)(3^2)=x^2-(i^2)(3^2)=x^2-(3i)^2\\\\i=\sqrt{-1}\to i^2=-1\\\\(*)=(x+5)[x^2-(3i)^2]=(x+5)(x-3i)(x+3i)\\\\Used:\ a^2-b^2=(a-b)(a+b)\\\\Answer:\ x^3+5x^2+9x+45=(x+5)(x-3i)(x+3i)$

4 0

3 years ago

Read 2 more answers

to raise money for her school field trip, Jane is selling candles. She has a large cancel for $8, and a small candle for $5. If

Fed [463]

Answer:

45 large candles
53 small candles

Step-by-step explanation:

Let c represent the number of large candles Jane sold. Then 98-c is the number of small candles. Her revenue is ...

8c +5(98-c) = 625

3c + 490 = 625 . . . . . . eliminate parentheses, collect terms

3c = 135 . . . . . . . . . . . . subtract 490

c = 45 . . . . . . . . . . . . . . divide by 3

Jane sold 45 large candles and 98-45 = 53 small candles.

7 0

3 years ago

Given the tables for f & g below, find the following: x f ( x ) g ( x ) 0 9 7 1 4 8 2 3 6 3 5 1 4 7 5 5 1 2 6 6 0 7 0 4 8 2

d1i1m1o1n [39]

$\\(i)\\
\text{We know that the average rate of change of a function f(x) on inerval [a,b] is}\\
\\
f_{avg}=\frac{f(b)-f(a)}{b-a}\\
\\
\text{so using this the average rate of chang of g for x on interval }[3,7] \text{ is}\\
\\
g_{avg}=\frac{g(7)-g(3)}{7-3}\\
\\
=\frac{4-1}{4}\\
\\
=\frac{3}{4}\\
\\
\text{so average rate of change of g on interval [3,7] is }\frac{3}{4}$

$\\(ii)\\ \text{Average rate of chang of f for x on interval }[3,7] \text{ is}\\
\\
f_{avg}=\frac{f(7)-f(3)}{7-3}\\
\\
=\frac{0-5}{4}\\
\\
=\frac{-5}{4}\\
\\
\text{so average rate of change of g on interval [3,7] is }\frac{-5}{4}$

$\\(iii)\\
\text{Since g(4)=5, so we have}\\
\\
f(g(4))=f(5)\\
\\
\text{since the value of f is 1 at x=5. so }\\
\\
f(g(4))=f(5)=1\\
\\
\text{hence we have, }f(g(4))=$ 1

$\\(iv)\\
\text{Since f(3)=5, so we have}\\
\\
g(f(3))=g(5)\\
\\
\text{since the value of g is 2 at x=5. so }\\
\\
g(f(3))=g(5)=2\\
\\
\text{hence we have, }g(f(3))=$ 2

5 0

3 years ago

2,6,10,.... find u16

Anni [7]

2000 to get 50 on u to be Nansen Lakshadweep. Hand

8 0

4 years ago

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