Mathematics, 29.09.2019 11:00 katswindle11

Right in equation of the ellipse the rest of the question is in the photo can you plz answer both

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QUESTION 1

The given ellipse has equation:

$4 {x}^{2} + 25 {y}^{2} = 100$

Divide through by 100 to get;

$\frac{ {x}^{2} }{25} + \frac{ {y}^{2} }{4} = 1$

The ellipse has it major axis on the x-axis.

${a}^{2} = 25$

$a = \pm5$

The vertices is

(-5,0) and (5,0).

Also

${b}^{2} = 4$

$b = \pm2$

The co-vertices are,

(0,-2), (0,2).

We use

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

${c}^{2} = {( \pm5)}^{2} - {( \pm2)}^{2}$

${c}^{2} = 25 - 4$

${c}^{2} =21$

$c = \pm \sqrt{21}$

The foci:

$(\pm \sqrt{21} ,0)$

We plot all these points and graph our ellipse.

See attachment

QUESTION 2.

If the ellipse has a vertex at

(0,-8) and a focus at (0,4) then

${a}^{2} = 64$

and

${c}^{2} = 16$

we can use

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

to determine the value if b.

${( - 8)}^{2} - {b}^{2} = {4}^{2}$

$64- {b}^{2} = 16$

${b}^{2} = 64 - 16$

${b}^{2} = 48$

The equation of the ellipse is

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

$\frac{ {x}^{2} }{48} + \frac{ {y}^{2} }{64} = 1$
Right in equation of the ellipse the rest of the question is in the photo can you plz answer both

Right in equation of the ellipse the rest of the question is in the photo can you plz answer both

Answer from: Quest

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

$a_{n}=a+(n-1)d\\ =19-5(n-1)\\=24-5n$

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

Answer from: Quest

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true

Bwill make the inequalities y < 2x + 1 and y> -x- 2 true. o a. region b o b. region c o c. r

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Right in equation of the ellipse the rest of the question is in the photo can you plz answer both