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3 years ago

F(x)=x^2+9 g(x)=x-9 find f(x)-g(x)

Mathematics

1 answer:

ololo11 [35]3 years ago

7 0

Answer:

x^2 - x + 18

Step-by-step explanation:

f(x) = x^2+9

g(x) = x-9

Therefore f(x)-g(x),

= x^2 + 9 - (x - 9)

= x^2 + 9 - x +9

<h2>= x^2 - x + 18</h2><h2 /><h2 /><h2><u><em>PLEASE MARK MY ANSWER AS BRAINLIEST!!!!!</em></u></h2>

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Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{8},0)(± 8 ,0)(, plus minus, square root o

Alchen [17]

Answer:

The equation of ellipse centered at the origin

$\frac{x^2}{18} +\frac{y^2}{10} =1$

Step-by-step explanation:

given the foci of ellipse (±√8,0) and c0-vertices are (0,±√10)

The foci are (-C,0) and (C ,0)

Given data (±√8,0)

the focus has x-coordinates so the focus is lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so c0-vertices are (0,±√10)

given focus C = ae = √8

Given co-vertices ( minor axis) (0,±b) = (0,±√10)

b= √10

The relation between the focus and semi major axes and semi minor axes are $c^2=a^2-b^2$

$a^{2} = c^{2} +b^{2}$

$a^{2} = (\sqrt{8} )^{2} +(\sqrt{10} )^{2}$

$a^{2} =18$

$a=\sqrt{18}$

The equation of ellipse formula

$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1$

we know that $a=\sqrt{18} and b=\sqrt{10}$

<u>Final answer:</u>-

<u>The equation of ellipse centered at the origin</u>

8 0

3 years ago

Read 2 more answers

Answer Question number 5 <br> Plzzzzzz

loris [4]

Answer:

Step-by-step explanation:

a = -4/3

d= -1 +4/3 =1/3

last term = 4 1/3 = 13/3

nth term = a +(n-1)d

$\frac{-4}{3}+(n-1)*\frac{1}{3}=\frac{13}{3}\\\\(n-1)*\frac{1}{3}=\frac{13}{3}-\frac{-4}{3}\\\\(n-1)*\frac{1}{3}=\frac{13+4}{3}\\\\(n-1)*\frac{1}{3}=\frac{17}{3}\\\\n-1=\frac{17}{3}*\frac{3}{1}\\\\n-1=17\\\\n=17+1=18$

Middle terms are 9the term and 10th term

$t_{9}=\frac{-4}{3}+8*\frac{1}{3}=\frac{-4}{3}+\frac{8}{3}=\frac{4}{3}\\\\t_{10}=\frac{-4}{3}+9*\frac{1}{3}=\frac{-4}{3}+\frac{9}{3}=\frac{5}{3}\\\\ t_{9}+t_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3\\\\t_{9}+t_{10}=3$