Algebra- Y=a(x-h)^2+k helppppp?

Question

question is ...prediction for when rocket would hit ground

equation is Y=a(x-h)^2+k

then i did

0=-5(x-7)^2+245

then v do

-5(x-7)^2=245

what do v do next?????

thankss u ppllll:)

thankss u ppllll:)

Answer 1

You made a mistake in your steps. This is what you should have.

Divide by -5.

-5(x-7)^2=-245
(x-7)^2 = 49

Square root both sides.

x-7 = 7

Add 7 to both sides.

x = 14.

Answer 2

Hello.
From what data you have given, I deduce that Y = 0, a= -5, h = 7 and k = 245.

You use the general formula (a - b)^2 = a^2 - 2ab + b^2. So you get
-5(x^2 - 14x + 49) + 245 = 0 which is eq. to -5 x^2 + 70 x -245 + 245 = 0. eq to -5 x^2 + 70 x = 0 eq. to -5x (x - 14) = 0.

At this point you can see that either -5x or (x - 14) needs to be 0. So either x = 0 or x = 14.

Hope this helps you.

Answer 3

I'm assuming you want to solve for x? Did -5, 7, and 245 come from an experiment?

If you want to solve for x, you should start from 0=-5(x-7)^2+245 and move 245 over to be -245 on the other side, so:

-245=-5(x-7)^2
(the sign is very important because you are about to take a square root, and you can't take a square root of a negative number.)

Then divide both sides by -5 to get:

49 = (x-7)^2

Then lake the square root of both sides:

7 = x-7

Then add 7

14 = x

Answer 4

Remember the rule FOIL.
First outside inside last.
rewrite the equation as:

-5((x-7)(x-7))=245
First do the FIRST two parts of the parenthesis (x*x)
Then the OUTSIDE parts (the -7*-7)
Then the INSIDE parts (-7*x)
Then the LAST parts (x*-7)

And add them all together giving your equation looking like

-5(x^2+49-7x-7x)=-245

So you now have

-5x^2+70x-245=-245

I guess that's -5x^2+70x=0

Divide it all by 5 to get rid of that.

so -x^2+14x=0

-x^2 = -14x
Divide both sides by -x
x=14

Hope you learned something and I hope I got it right...

Answer 5

it should be -245 on the right side, then divide by -5 both sides

(x-7)^2=49
take square root of both sides
x-7=7
x=14

Answer 6

divide both sides by -5...

[-5(x-7)^2]/5=[245]/5

that cleans it up a bit to:
(x-7)^2 = 49

Then you take the square root of both sides...
The square root in effect 'cancels out' the ^2.

Just don't forget to take the square root of both sides of the equation. What you do to one sid eyou HAVE to do to the other side.

then take care of the 7 by adding it to both sides...

Answer 7

0= -5(x-7)^2+245
5(x-7)^2=245
(x-7)^2=49
taking sqroots,
i.e either x-7=7 or x-7=-7
that gives x=14 or x=0
value of x depends upon what it is refering to, as a result one value shuld be ignored.

Answer 8

i do no longer think of you typed out the question properly... If the two preliminary expenses are the comparable (40 seven.ninety 8), Plan B will in no way value under Plan A no count what number miles you force. in case you force 0 miles the fee often is the comparable, yet any mileage by any capacity capacity the 2d plan will continually value extra. in many cases, those questions have a bigger base value plus smaller according to mile value whilst in comparison with a decrease base value yet a bigger according to mile value. My guess is to make sure the question returned because of the fact the 2d plan could desire to have a plenty decrease commencing value... The equation could in all probability land up looking like this: (value for 2 days plan A) + 0.13X = (value for 2 days plan B) + 0.15X clean up for X.

Answer 9

-5(x-7)² = -245 ........... you dropped your minus sign
(x-7)² = 49
x - 7 = ±7
x = 7 ± 7
x = 7 - 7 = 0 or
x = 7 + 7 = 14

Answer 10

CAN BE REPRESENTED AS:

0 = - 5 (x - -7)^2 + 245

SIMPLIFIED TO:

0 = - 5·x·(x + 14)

EXPANDED TO:

0 = x^2 + 14x

x^2 = -14x

(x^2 = -14x)/-14x

-(x/14) = 1

-x = -14

x = 14