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3 years ago

I can't figure it out. 2x+2y=2-4x+4y=12

Mathematics

1 answer:

koban [17]3 years ago

6 0

Answer:

how to solve 2x+2y=2 , -4x+4y=12

this is solving equations using substitution .... i have ALOT more but i juss dnt get this stuff

2x + 2y = 2 -4x + 4y = 12

x + y = 1

x = 1 - y

-4x + 4y = 12

-4(1 - y) + 4y = 12

(-4 + 4y) + 4y = 12

-4 + 8y = 12

8y = 16

y = 2

2x + 2y = 2

2x + 2(2) = 2

2x + 4 = 2

2x = -2

x = -1

Check.

-4x + 4y = 12

-4(-1) + 4(2) = 12

4 + 8 = 12

Answer: x = -1 and y = 2

Step-by-step explanation:

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(a) The product of $(2x-4)\ and\ (3x^2-x+4)$ is calculated to be $=6x^3-14x^2+12x-16$

(b)The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

Step-by-step explanation:

(a)

$(2x-4)(3x^2-x+4)\\\\=(2x.3x^2)+(2x.-x)+(2x.4)+(-4.3x^2)+(-4.-x)+(-4.4)\\\\=6x^3-2x^2+8x-12x^2+4x-16\\\\=6x^3-14x^2+12x-16$

$(4x-2)(3x^2-x+4)\\\\=(4x.3x^2)+(4x.-x)+(4x.4)+(-2.3x^2)+(-2.-x)+(-2.4)\\\\=12x^3-4x^2+16x-6x^2+2x-8\\\\=12x^3-10x^2+18x-8$

The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

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3 years ago

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

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Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for h:

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for h.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

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3 years ago

I can't figure it out. 2x+2y=2-4x+4y=12​

I can't figure it out. 2x+2y=2-4x+4y=12