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lilavasa [31]
3 years ago
7

-33 is greater than x .​

Mathematics
2 answers:
IRINA_888 [86]3 years ago
8 0

Answer:

-33>x

Step-by-step explanation:

algol [13]3 years ago
3 0
The format would be “-33>x”
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Could someone help me out?
Studentka2010 [4]

The equation of the line is y = -11x + 232

<h3>How to determine the equation?</h3>

The given parameters are:

Slope (m)= -11

Point (x1, y1) = (31, -109)

The linear equation is then calculated as:

y = m(x - x1) + y1

This gives

y = -11(x - 31) - 109

Evaluate the product

y = -11x + 341 - 109

Evaluate the like terms

y = -11x + 232

Hence, the equation of the line is y = -11x + 232

Read more about linear equations at:

brainly.com/question/14323743

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3 0
2 years ago
What is the exponential growth formula?
maksim [4K]

Answer:

To calculate exponential growth, use the formula y(t) = a__ekt, where a is the value at the start, k is the rate of growth or decay, t is time and y(t) is the population's value at time t.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Solve for x. 9^6 = x^3 • 9^3
Savatey [412]

Answer:

X=56/(27*3)

hope it helps you

6 0
2 years ago
A high school will graduate 100 students. of these, 52 students plan to attend college.
marishachu [46]
A total of 48 students will not go to college
8 0
3 years ago
An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d
denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. w\times 2w\times h=36

h=\frac{18}{w^2}

The equation form when top is open,

f(w)=2w^2+2wh+2(2w)h

Substitute the value of h,

f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})

f(w)=2w^2+\frac{36}{w}+\frac{72}{w}

f(w)=2w^2+\frac{108}{w}

Derivate w.r.t 'w',

f'(w)=4w-\frac{108}{w^2}

For critical point put it to zero,

4w-\frac{108}{w^2}=0

4w=\frac{108}{w^2}

w^3=27

w^3=3^3

w=3

Derivate the function again w.r.t 'w',

f''(w)=4+\frac{216}{w^3}

For w=3, f''(3)=4+\frac{216}{3^3}=12 >0

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = \frac{18}{3^2}=2\ ft

Therefore, the dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

4 0
3 years ago
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