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3 years ago

Need help on this hopefully y'all can help me on it

Mathematics

2 answers:

Travka [436]3 years ago

4 0

The formula for finding the volume of a cylinder is V=πr²h where V is the volume, π is 22/7, r is the radius and, h is the height or V=1/4πd²h where V is the volume,π=22/7 ,d is the diameter and h is the height.
Given that the diameter is 12 inches and the height is 10 inches, the volume is
V=1/4πd²h
V=1/4*22/7*12*12*10
V=1131 inches³(rounded to nearest ones or unit)
Therefore the volume of the cylindrical container is approximately 1131cm³

exis [7]3 years ago

3 0

Use the formula. Change diameter to radius. Fill in formula and calculate.

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Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3

NikAS [45]

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. $f(x)=-2x+1$

If we compare this equation with the slope intercept form of a line which is given as

$y=mx+c$

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

$y=-2(1)+1$

$y=-2+1 = -1$

Let us find another point by putting x = 2 and solving it for y

$y=-2(2)+1$

$y=-4+1 = -3$

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

$y=-2(-2)+1$

$y=+1 = 5$

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) , (2,-3) and (-2,5) we joint them on line to obtain our line

$g(x)=y=x^2-2x-3$

$y=x^2-2x+1-1-3$

$y=(x-1)^2-4$

$(y+4)=(x-1)^2$

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; $y=y=(0)^2-2(0)-3=0-0-3=-3$ ; (0,-3)

ii)x=-1 ; $y=(-1)^2-2(-1)-3=1+2-3=0$ ; (-1,0)

iii) x=2 ; $y=(2)^2-2(2)-3 = 4-4-3 =-3$ ;(2,-3)

iii) x=1 ; $y=(1)^2-2(1)-3 = 1-2-3 =-4$ ;(1,-4)

iii) x=-2 ; $y=(-2)^2-2(-2)-3 = 4+4-3 =5$ ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.