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4 years ago

What’s the answer to #12? and why

Mathematics

2 answers:

Elena L [17]4 years ago

6 0

Remark
If there are 5 distinct zeros that means either that the x axis is crossed the x axis 5 different places or touched the x axis in 1 place out the 5. Touching in one place means that an even number of roots are the same.

So let's go through all of them to get an answer of 5.

A has 4 x intercepts. It is not the right answer. We need 5.
B has 4 x intercepts. It is not the right answer. We need 5.
C has 6 x intercepts. Not the one we want.
D has 5 x distinct zeros. The wording is a bit tricky. It does not matter than one of them just touches the x axis. There could be an even number of distinct zeros there, but it only counts as one root.

An example of such a graph is f(x)= $\left(\frac{1}{10}(x+2)(x+1.5)(x+1)(x-2)^4(x-3)\right)$

Answer D <<<<<

expeople1 [14]4 years ago

3 0

It is D. Hope it's right

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Step-by-step explanation:

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3 years ago

I need to find the surface area and volume of all three figures. If you could provide what equations you used to I will be grate

Margarita [4]

Answer:

Part 1) Sphere The surface area is equal to $SA=196\pi\ m^{2}$ and the volume is equal to $V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) Cone The surface area is equal to $SA=(16+4\sqrt{65})\pi\ units^{2}$ and the volume is equal to $V=\frac{112}{3}\pi\ units^{3}$

Part 3) Triangular Prism The surface area is equal to $SA=51.57\ mm^{2}$ and the volume is equal to $V=17.388\ mm^{3}$

Step-by-step explanation:

Part 1) The figure is a sphere

a) Find the surface area

The surface area of the sphere is equal to

$SA=4\pi r^{2}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$SA=4\pi (7)^{2}$

$SA=196\pi\ m^{2}$

b) Find the volume

The volume of the sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$V=\frac{4}{3}\pi (7)^{3}$

$V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) The figure is a cone

a) Find the surface area

The surface area of a cone is equal to

$SA=\pi r^{2} +\pi rl$

we have

$r=4\ units$

$h=7\ units$

Applying Pythagoras Theorem find the value of l (slant height)

$l^{2}=r^{2} +h^{2}$

substitute the values

$l^{2}=4^{2} +7^{2}$

$l^{2}=65$

$l=\sqrt{65}\ units$

$SA=\pi (4)^{2} +\pi (4)(\sqrt{65})$

$SA=16\pi +4\sqrt{65}\pi$

$SA=(16+4\sqrt{65})\pi\ units^{2}$

b) Find the volume

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=4\ units$

$h=7\ units$

substitute

$V=\frac{1}{3}\pi (4)^{2}(7)$

$V=\frac{112}{3}\pi\ units^{3}$

Part 3) The figure is a triangular prism

a) The surface area of the triangular prism is equal to

$SA=2B+PL$

where

B is the area of the triangular base

P is the perimeter of the triangular base

L is the length of the prism

Find the area of the base B

$B=\frac{1}{2} (2.7)(2.3)=3.105\ mm^{2}$

Find the perimeter of the base P

$P=2.7*3=8.1\ mm$

we have

$L=5.6\ mm$

substitute the values

$SA=2(3.105)+(8.1)(5.6)=51.57\ mm^{2}$

b) Find the volume

The volume of the triangular prism is equal to

$V=BL$

where

B is the area of the triangular base

L is the length of the prism

we have

$B=3.105\ mm^{2}$

$L=5.6\ mm$

substitute

$V=(3.105)(5.6)=17.388\ mm^{3}$

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