Find an equation for the line that passes through the points (5, -4) and (1, -2)

What is the midpoint

mrs_skeptik [129]

The answer is (4, 3). Hope this helps!

8 0

3 years ago

Solve this system by substitution (3 variables) brainliest answer goes to first answer and 8 points!

Natalka [10]

Answer:Solve this system by substitution (3 variables) brainliest answer goes to first answer and 8 points!

-x -y -z = -8

-4x +4y +5z = 7

2x+2z=4

Show steps please!

Step-by-step explanation: a-z=3-2=657u568959jb5j656856 thats the answer your welcome

4 0

3 years ago

I WILL AWARD BRAINLIEST TO THE FIRST PERSON TO ANSWER HONESTLY

ruslelena [56]

Answer:

1,3,5

Step-by-step explanation:

Your first answer is correct,the third one too,and your fifth too.

Hope this helps:)

Pls mark brainlist

3 0

3 years ago

How do you integrate <br><img src="https://tex.z-dn.net/?f=%283x%5E%7B2%7D%20%20%2B%203%29%20%5E%7B3%7D%20" id="TexFormula1" tit

Phoenix [80]

Answer:

Since i don't know what formula is mentioned. I just used the easiest way to solve :)

Step-by-step explanation:

$(3x^2 + 3)^3 = (3x^2)^3 + 3^3 + 3((3x^2)^2)(3) + 3(3x^2)(3^2)$

$= 27x^6 + 27 + 3(9x^4 \times 3) + 3(3x^2 \times 9)\\\\=27x^6 + 27 + 81x^4 + 81x^2\\\\=27x^6 + 81x^4 + 81x^2 + 27\\\\$

$\int\limits {(3x^2+3)^3} \, dx = \int\limits {27x^6 + 81x^4+81x^2 + 27} \, dx$

$= \frac{27}{7}x^7 + \frac{81}{5}x^5+\frac{81}{3}x^3 + 27x +C\\\\= \frac{27}{7}x^7 + \frac{81}{5}x^5+27x^3 + 27x +C\\\\$

8 0

3 years ago

Please help! ive been stuck on this for so long and i just keep getting frustrated can someone help walkme through this?

bagirrra123 [75]

For firework launched from height 100ft with initial velocity 150ft/sec, equation made is correct

(a) equation will be $h(t) = -16t^2+150t+100$

(b) Now we have to see when it will land. At land or ground level height h will be equal to 0. So simply plug 0 in h place in equation made in part (a)

$0 = -16t^2 + 150t + 100$

Now we have to solve this quadratic. We will use quadratic formula method to solve this equation.

$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

a = -16, b = 150, c = 100.

Plugging these values in quadratic formula we get

$t = \frac{-150 \pm \sqrt{150^2-4(-16)(100)}}{2(-16)}$

$t = \frac{-150 \pm \sqrt{22500+6400}}{-32}$

$t = \frac{-150 \pm \sqrt{28900}}{-32}$

$t = \frac{-150+170}{-32} = \frac{20}{-32} = -0.625$

time cannot be negative so we will drop this answer

then $t = \frac{-150-170}{-32} = \frac{-320}{-32} = 10$

So 10 seconds is the answer for this

(c) To make table simply plug various value for t like t =0, 2, 4, 6, 8 till 10. Plug values in equation mad in part (a) and find h value for each t as shown

For t =0 seconds, $h = -16(0)^2+150(0)+100 = 100 feet$

For t =2 seconds, $h = -16(2)^2+150(2)+100 =336 feet$

For t =4 seconds, $h = -16(4)^2+150(4)+100 = 444 feet$

For t =6 seconds, $h = -16(6)^2+150(6)+100 = 424 feet$

For t =8 seconds, $h = -16(8)^2+150(8)+100 = 276 feet$

For t =10 seconds, $h = -16(10)^2+150(10)+100 = 0 feet$

(d) Axis of symmetry is given by formula

$x = \frac{-b}{2a}$

$t = \frac{-150}{2(-16)} =\frac{-150}{-32} = 4.6875$

t = 4.6875 is axis of symmetry line

(e) x-coordinate of vertex is again given by formula

$x = \frac{-b}{2a}$

so t = 4.6875

then to find y coordinate we will plug this value of t as 4.6875 in equation made in part (a)

For t =4.6875, $h = -16(4.6875)^2+150(4.6875)+100 = 451.563$

so vertex is at (4.6875, 451.563)

(f) As the firework is launched so in starting time is t=0, we cannot have time before t=0 (negative values) practically. Also we cannnot have firework going down into the ground so we cannot have h value negative physically.

6 0

3 years ago