Factor the expression: 2x^2 + 5x + 2

Help me with geometry pleasee ( Radom answers will be reported )

IrinaVladis [17]

A. jgb=jb+Eh/2
35+45=80/2=40 JGB =40°
B. since it is a line HGJ =180-40
HGJ= 140°
C. tangents to circles have equal lengths if they intercept the way gd and gf do eachother
set the lengths equal to themselves
4x+5=6x-8
-4x both sides
5=2x-8
+8 both sides
13=2x
÷2 both sides
x=13/2 or 6.5
Plug in
4 (6.5)+5=
26+5=31ft

8 0

3 years ago

How do I find the volume of this?

lorasvet [3.4K]

Answer:

(64+320π) cm^3

Step-by-step explanation:

The volume of a cube is the length*width*height, so

the volume of this cube is 4*4*4=64 cm^3

The volume of a cylinder is the base*height

The base of a cylinder is the area of a circle, which is π*the radius of the circle squared

The diameter of the circle base is 16, and the radius is half the diameter, so it is 16/2=8.

The area of the base of the cylinder is π8^2=64π

Now, multiply the base by the height, 5

64π*5=320πcm^3

We now have the volume of the cube and the volume of the cylinder, so now we just add them:

(64+320π) cm^3

7 0

3 years ago

The equation below represents Function A and the graph represents Function B:

kozerog [31]

The slope of function A is the coefficient of x, which is 6.

The slope of function B is the rise divided by the run, which is 3/1 = 3.

The appropriate choice is ...

... c) Slope of Function A = 2×Slope of Function B

8 0

4 years ago

Please help me with this one, I’m really stuck on it.

ANTONII [103]

I don’t know if this will help but here you go

6 0

3 years ago

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

8 0

3 years ago