14 points to brainliest

Multiple choice question please answer it for me

timofeeve [1]

The answer is the second choice, or B; 240 ft^2.

You have to find the area of all four triangles and just add those all together.

Two of the triangles have an area of
5*16=80
Multiply that by two
80*2=160

The other two smaller triangles have an area of
8*5=40
Multiply that by 2 also
40*2=80

Add 160+80 and the answer is 240.

6 0

3 years ago

3x^2 + 4x = -8<br> solve using quadratic formula

Firdavs [7]

Use the quadratic equation to solve this (image of the quadratic equation is below)

Remember that quadratic functions are set up like so:

$ax^{2} +bx +c = 0$

To make the equation 3x² + 4x = -8 into a quadratic function you must bring -8 to the left side of the equation so it equals zero. To do this add 8 to both sides

3x² + 4x + 8= -8 + 8

3x² + 4x + 8 = 0

That means that in this equation...

a = 3

b = 4

c = 8

^^^Plug these numbers into the quadratic equation and solve (Keep in mind that +/- is ± )

$\frac{-4+/-\sqrt{(4)^{2}-4(3)(8)}}{2(3)}$

$\frac{-4+/-\sqrt{16-96}}{6}$

$\frac{-4+/-\sqrt{-80}}{6}$

^^^There is no real solution to this equation. You must have an imaginary solution

$\frac{-4+/-i\sqrt{80}}{6}$

Simplify

$\frac{-2+/-2i \sqrt{5}}{3}$

Hope this helped!

~Just a girl in love with Shawn Mendes

5 0

2 years ago

I need help with the last two

Luba_88 [7]

.07 is what I believe the answer is

3 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 25

Assoli18 [71]

Answer:

(a) The velocity after 2 second is 5.9 m/s

The velocity after 4 second is -13.7 m/s.

(b) The projectile reaches its maximum height after 2.60 s of projection.

(c)The maximum height that is attained by the projectile is 37.18 m.

(d)Therefore the projectile hits the ground after 5.36 seconds of projection.

(e)The velocity of the projectile when it hits the ground is 27.03 m/s

Step-by-step explanation:

Given that, a projectile shot vertically upward from a point 4 m above the ground with a initial velocity of 25.5 m/s.

The height of the projectile after t seconds is

$h=4+25.5t-4.9t^2$

where h is in meter.

(a)

We use the formula

v=u+at

V= final velocity

u = initial velocity = 25.5 m/s

a = acceleration= acceleration due to gravity= 9.8 m/s²

Since the object moves upward direction and acceleration due to gravity is downward direction. So here a= -9.8 m/s.

v(2)= 25.5+(-9.8)×2

=25.5-19.6

=5.9 m/s

And when t= 4

v(4)= 25.5+(-9.8)×4

=25.5-39.2

= -13.7 m/s

The velocity after 2 second is 5.9 m/s

The velocity after 4 second is -13.7 m/s.

(b)

At its maximum height,the velocity of the projectile is zero. i.e v=0

∴0=25.5+(-9.8)t

⇒9.8t=25.5

$\Rightarrow t=\frac{25.5}{9.8}$

⇒t = 2.60 s

The projectile reaches its maximum height 2.60 s after projection.

(c)

To find the maximum height, we are putting t= 2.60 in this equation $h=4+25.5t-4.9t^2$ .

$\therefore h= 4+(25.5\times 2.60)-(4.9\times 2.60^2)$

=37.18 m

The maximum height that is attained by the projectile is 37.18 m.

(d)

When the projectile hits the ground the height will be zero i.e h=0

From the equation of height we get

$\therefore h=0=4+25.5t-4.9t^2$

$\Rightarrow 4+25.5t-4.9t^2=0$

$\Rightarrow t=\frac{-25.5\pm\sqrt{25.5^2-4(-4.9).4}}{2(-4.9)}$

⇒t= -0.15 ,5.36

Therefore it hits the ground after 5.36 seconds of projection.

(e)

To find the velocity we use the formula v=u+at

Here v = final velocity=?

u=25.5 m/s,

t = 5.36 s

a= -9.8m/s²

v=25.5+(-9.8)5.36

= -27.03 m/s

Negative sign denoted that the motion of the projectile is downward direction.

The velocity of the projectile when it hits the ground is 27.03 m/s.

6 0

3 years ago

SOMEONE PLZ HELP ME!!!

Natali [406]

Answer:

The answer to this question is B

Step-by-step explanation: By simplifying both equations to a x = ? both equations lead up to x = -7. Inversely, you can multiply both sides of the first equation by -3 and get the same equation as the second equation.

6 0

3 years ago