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3 years ago

PLZ SHOW ALL WORK

1 answer:

3 0

Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width: w = 12 - 2x
length: l = 18 - 2x
height: h = x

Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x

Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6

Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160

Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224

And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.

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What is the value of x 1/5×-2/3y=30 when y=15

Alinara [238K]

Equation.
1/5 x - 2/3 y = 30

y= 15

Substitution.
1/5 x - 2/3 (15) = 30

1/5 x - 10 = 30

1/5 x = 30 + 10

1/5 x = 40

x = 200.

3 0

3 years ago

What is the type of conic section given by the equation x^2 - 9y^2 = 900 and what is the domain and range?

Valentin [98]

A conic section is a curve obtained by the intersection of the surface of a cone with a plane. A conic section can be a circle, a hyperbola, a parabola, and an ellipse.

For a circle, the general equation of a circle with center, (a, b), and a radius, r, is of the form
$(x-a)^2+(y-b)^2=r^2$

For a hyperbola, the general equation of a hyperbola with center (h, k), and a and b half the lengths of the major and the minor axis respectively is of the form.
$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} =1$

For a parabola, the general equation of a parabola with center (h, k), and a multiplier a is of the form
$y-k=a(x-h)^2$

For an ellipse, the general equation of an ellipse with center (h, k), and a and b half the lengths of the major and the minor axis respectively.
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} =1$

Given the equation
$x^2-9y^2=900$
It can be rewritten as
$\frac{(x-0)^2}{900} - \frac{(y-0)^2}{100} =1 \\ \\ \frac{(x-0)^2}{30^2} - \frac{(y-0)^2}{10^2} =1$
This gives an equation of a hyperbola with center (0, 0), half the length of the major axis = 30 and half the length of the minor segment = 10.

The domain of the equation is all real values of x.

8 0

3 years ago

Algebra, please help. 2 phone companies prices.

nordsb [41]

Answer:

At 10 minutes the phone call costs the same for both companies. Hopefully this was helpful. :)

Step-by-step explanation:

Create equation:

f(x)=3x+35

f(x)=2x+45

x=minutes

Solve:

Company A: Company B:

0 35 45

1 38 47

2 41 49

3 44 51

4 47 53

5 50 55

6 53 57

7 56 59

8 59 61

9 62 63

10 65 65

3 0

3 years ago

Two consecutive integers have a sum of 73. Find the integers.

patriot [66]

Here's how to do it using algebra:

Let the smaller integer be 'x' . Then the larger integer is (x+1).

Their sum is (x) + (x+1).

That's (2x + 1) .

The problem says their sum is 73, so we can write . . . 2x + 1 = 73 .

Subtract 1 from each side . . . 2x = 72

Divide each side by 2 . . . x = 36

The smaller integer is 36 .

The larger integer is (x+1) . That would be 37.

The two consecutive integers are 36 and 37 .

_______________________________________

Here's how to do it using your brain:

-- If they were both the same number, then their sum would be double it, and each integer would be half of 73 .

-- That's 36.5 .

-- But 36.5 is not even an integer, so neither integer can be 36.5. Is there something we can do to a pair of 36.5s to make them whole numbers (integers), make them different, and make them consecutive, but not let their sum change ?

-- Well, if we slice off a piece from one of them and glue it onto the other one, then we'll wind up with two numbers that are different, and their sum won't change.

-- What happens if we take the .5 off of one of them and glue it onto the other one ? Then one of the 36.5s shrinks to become 36, and the other one grows to become 37 .

-- Well whaddaya know ! Now we have two numbers that are consecutive integers, and they add up to 73 . Great move !

4 0

3 years ago

Read 2 more answers

Which ordered pair is the solution to the system of linear equations y=-7X+2 and y = 9%-14?

Dmitriy789 [7]

Answer:

y = -7x + 2

y = 9x - 14

-7x + 2 = 9x - 14

14 + 2 = 9x + 7x

16 = 16x

1 = x

y = -7x + 2

y = -7(1) + 2

y = -7 + 2

y = -5

solution is : (1,-5)

7 0

3 years ago