Ask question

Ask question

All categories

3 years ago

6

Which is the focus of a parabola with equation y2=-12x

1 answer:

Morgarella [4.7K]3 years ago

3 0

We have
y²=-12x

the vertex is the point (0,0)------------- > see the attached figure
<span>
standard form: (y-k)^2=4p(x-h), with (h,k) being the (x,y) coordinates of the vertex.
vertex(0,0)
4p=-12---- > p=-12/4
p=-3
Focus:(-3,0)

</span>the answer is (-3,0)

You might be interested in

Which one weighs more 500 pounds or 200 kilograms

Katyanochek1 [597]

1 Pound is 0.453592 Kilograms
500 Pounds is 226.796 Kilograms
Hope i helped!
Please Leave a thanks! Thanks!

5 0

3 years ago

Read 2 more answers

Need some help with these problems please.

goldenfox [79]

Answer:

See below.

Step-by-step explanation:

a.

Divide the leading terms:

6x^4 / 2x^2 = 3x^2

3x2 is a parabola so the long run is

as x ---> +/- infinity r(x) ----> + infinity. (answer).

b.

10,000x^3 / 50 x^3

= 200.

So r(x) as a horizontal asymptote at y ( r(x)) = 200. (answer).

3 0

3 years ago

Please help me answer for 35 and 36

dexar [7]

Answer:

35. $f(2)=16$

36. $f(2)=77$

Step-by-step explanation:

Question 35:

Given:

$f(1)=32$

$f(n)=f(n-1)\times \frac{1}{2}$

Plug in 2 for $n$ and simplify. This gives,

$f(2)=f(2-1)\times \frac{1}{2}\\f(2)=f(1)\times \frac{1}{2}\\f(2)=32\times \frac{1}{2}\\f(2)=16$

Therefore, term 2 is 16.

Question 36:

Given:

$f(1)=11$

$f(n)=f(n-1)\times 7$

Plug in 2 for $n$ and simplify. This gives,

$f(2)=f(2-1)\times 7\\f(2)=f(1)\times 7\\f(2)=11\times 7\\f(2)=77$

Therefore, term 2 is 77.

8 0

3 years ago

Factorize 2a^2+7a-15

loris [4]

$\huge\bf \pink {\underline{Solution :-}}$

$2 {a}^{2} + 7a - 15$

$= 2 {a}^{2} + 10a - 3a - 15$

$= 2a(a + 5) - 3(a + 5)$

$= (a + 5)(2a - 3)$

$\sf \red{Hence, Answer \: is \: (a + 5)(2a - 3).}$

$\\$

$\bf \purple{ \underline{ Important \: Formulas \: for \: Factorization :-}}$

$• \: {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2}$

$• \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

$• \: {a}^{2} - 2 ab + {b}^{2} = {(a - b)}^{2}$

$• \: {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3} = {(a + b)}^{3}$

$• \: {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} = {(a - b)}^{3}$

$• \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

$• \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$• \: (a + b)( {a}^{2} -ab + {b}^{2} ) = {a}^{3} + {b}^{3}$

$• \: (a - b)( {a}^{2} + ab + {b}^{2} ) = {a}^{3} - {b}^{3}$

$• \: {( \frac{a + b}{2} )}^{2} - ( {\frac{a - b}{2} } )^{2} = ab$

5 0

3 years ago

Read 2 more answers

Answer to be marked as brainlist

zimovet [89]

Answer:

The SF is 3

Step-by-step explanation:

The sides are 3x the length of the sides of ∆ABC

4 0

2 years ago