Help please help me please

PLEASE HELP ASAP -- WITH THE WORKK<br><br>50 points

MAXImum [283]

Part-A:-

$\\ \sf\longmapsto x^2+3x-70(4x-12)$

$\\ \sf\longmapsto 4x^2+12x-280-12x^2+36x+840$

$\\ \sf\longmapsto -8x^2+48x+560$

Now

$\\ \sf\longmapsto (f.g)(3)$

$\\ \sf\longmapsto -8(3)^2+48(3)+560$

$\\ \sf\longmapsto -72+144+560$

$\\ \sf\longmapsto 72+560$

$\\ \sf\longmapsto 632$

Part-B:-

$\\ \sf\longmapsto f(3)$

$\\ \sf\longmapsto (3)^2+3(3)-70$

$\\ \sf\longmapsto 9+9-70$

$\\ \sf\longmapsto 18-70$

$\\ \sf\longmapsto -52$

Now

$\\ \sf\longmapsto g(f(3))$

$\\ \sf\longmapsto g(-52)$

$\\ \sf\longmapsto 4(-52-3)$

$\\ \sf\longmapsto 4(-55)$

$\\ \sf\longmapsto -220$

Part:-C

$\\ \sf\longmapsto g(-2)$

$\\ \sf\longmapsto 4(-2-3)$

$\\ \sf\longmapsto 4(-5)$

$\\ \sf\longmapsto -20$

Now

$\\ \sf\longmapsto f(g(-2))$

$\\ \sf\longmapsto f(-20)$

$\\ \sf\longmapsto (-20)^2+3(-20)-70$

$\\ \sf\longmapsto 400-60-70$

$\\ \sf\longmapsto 400-130$

$\\ \sf\longmapsto 270$

Part-D:-

g×f=f×g

$\\ \sf\longmapsto (g.f)(-2)$

$\\ \sf\longmapsto -8(-2)^2+48(-2)+560$

$\\ \sf\longmapsto 32-96+560$

$\\ \sf\longmapsto -64+560$

$\\ \sf\longmapsto 496$

6 0

2 years ago

PLS HELP ASAP ILL GIVE BRAINLKEST THANKS 10 POINTS

KATRIN_1 [288]

1st question

Answer: y = 200x

y = 400

x = 2

400 = 200*2

2nd question

Answer: y = 2x

y = 6

x = 3

6 = 2*3

8 0

2 years ago

What is the volume of the composite figure? round to the nearest hundredth

Mariana [72]

Answer:

Composite figures are composed of several geometric shapes and are three-dimensional shapes. To find the volume of the entire shape you find the volume of each individual shape and add them together. The second figure consists of a cylinder and a hemisphere

5 0

2 years ago

N-4=16 solve equation

gizmo_the_mogwai [7]

n - 4 = 16 Given

+ 4 + 4 Cancel out 4 to isolate n variable

n = 20

Prove.

20 - 4 = 16

16 = 16

True.

6 0

2 years ago

Read 2 more answers

A circle is centered at J(3, 3) and has a radius of 12.

stealth61 [152]

Answer:

$(-6,\, -5)$ is outside the circle of radius of $12$ centered at $(3,\, 3)$ .

Step-by-step explanation:

Let $J$ and $r$ denote the center and the radius of this circle, respectively. Let $F$ be a point in the plane.

Let $d(J,\, F)$ denote the Euclidean distance between point $J$ and point $F$ .

In other words, if $J$ is at $(x_j,\, y_j)$ while $F$ is at $(x_f,\, y_f)$ , then $\displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}$ .

Point $F$ would be inside this circle if $d(J,\, F) < r$ . (In other words, the distance between $F\!$ and the center of this circle is smaller than the radius of this circle.)

Point $F$ would be on this circle if $d(J,\, F) = r$ . (In other words, the distance between $F\!$ and the center of this circle is exactly equal to the radius of this circle.)

Point $F$ would be outside this circle if $d(J,\, F) > r$ . (In other words, the distance between $F\!$ and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between $J$ and $F$ :

$\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145} \end{aligned}$ .

On the other hand, notice that the radius of this circle, $r = 12 = \sqrt{144}$ , is smaller than $d(J,\, F)$ . Therefore, point $F$ would be outside this circle.

5 0

3 years ago