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Goshia [24]
4 years ago
5

I will mark brainliest!

Mathematics
1 answer:
gayaneshka [121]4 years ago
7 0
= 4x - 11

x is the input, y is the output.

So if you want to know what input is needed to get an output of 10, set y to 10 and solve for x:

10 = 4x - 11
21 = 4x
x = 21/4
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For the function f(x)=x^2-5x+6, where is the y-intercept?
Jlenok [28]
I believe the answer is 6
5 0
3 years ago
A cylinder has a height h, and a radius (3h-2)
tekilochka [14]

Answer:

V= 4\pi (9h^2 -12h + 4)

Step-by-step explanation:

Given

Shape: Cylinder

Radius = 3h - 2

Height = h

Base\ Area = \pi (3h - 2)^2

Required

Simplify the volume

The volume is calculated as:

Volume = Base\ Area * Height

Substitute values for Base Area and Height

V= \pi(3h-2)^2*h

Expand the bracket

V= \pi(3h-2)(3h-2)*h

Open brackets

V= \pi (9h^2 -6h - 6h + 4) *h

V= \pi (9h^2 -12h + 4) *h

V= 4\pi (9h^2 -12h + 4)

6 0
3 years ago
A rectangular area is to be fenced in with 300 feet of chicken wire. find the maximum area that can be enclosed.
GarryVolchara [31]
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.

2(x+y) = 300
x+y = 150
y = 150-x

A=x(150-x) <--(substitution)

The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150

So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.

A=75(150-75)
A=75*75
A=5625

So the maximum area that can be enclosed is 5625 square feet.
6 0
3 years ago
Write the equation of the line in fully simplified slope-intercept form.
Alona [7]
Y=3/1x+-1 hope this is correct
5 0
3 years ago
Why does a radical function with an even index only appear on one sode of the x-axis while a radical woth an odd index appears o
Vladimir79 [104]

Consider the equation y = x^2. No matter what x happens to be, the result y will never be negative even if x is negative. Example: x = -3 leads to y = x^2 = (-3)^2 = 9 which is positive.

Since y is never negative, this means the inverse x = sqrt(y) has the right hand side never be negative. The entire curve of sqrt(x) is above the x axis except for the x intercept of course. Put another way, we cannot plug in a negative input into the square root function for this reason. This similar idea applies to any even index such as fourth roots or sixth roots.

Meanwhile, odd roots such as a cube root has its range extend from negative infinity to positive infinity. Why? Because y = x^3 can have a negative output. Going back to x = -3 we get y = x^3 = (-3)^3 = -27. So we can plug a negative value into the cube root to get some negative output. We can get any output we want, negative or positive. So the range of any radical with an odd index is effectively the set of all real numbers. Visually this produces graphs that have parts on both sides of the x axis.

3 0
3 years ago
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