All categories

3 years ago

Solve using elimination -3-X=0 -3y=-3-2x

Mathematics

1 answer:

ryzh [129]3 years ago

7 0

<h2>The solution is (x, y) = (-3, -1)</h2>

<em><u>Solution:</u></em>

<em><u>Given, system of equations are:</u></em>

-3 - x = 0 --------- eqn 1

-3y = -3 - 2x -------- eqn 2

From eqn 1,

- 3 - x = 0

<em><u>Substitute x = -3 in eqn 2</u></em>

-3y = - 3 - 2(-3)

-3y = -3 + 6

-3y = 3

Thus the solution is (x, y) = (-3, -1)

You might be interested in

What’s the answer to this one! 15 points!

Phoenix [80]

Answer:

Step-by-step explanation:

Because they are opposite of each other

5 0

3 years ago

Read 2 more answers

16.4 divide 0.8 what is the answer

fgiga [73]

20.5 is the answer, have a great day!

5 0

2 years ago

Read 2 more answers

Which is the simplified form of the expression 3(7/6x + 4) - 2(3/2-5/4x)<br><br><br>

mars1129 [50]

Answer:

Step-by-step explanation:

8 0

3 years ago

1. Find the simple interest on a $3,219.00 principal, deposited for six years at a rate of 1.51%.

Nataliya [291]

Answer:

bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

4 0

3 years ago

Math question

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

$h = \frac{V}{\pi\cdot r^{2}}$

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

$h = \frac{V}{\pi\cdot r^{2}}$

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

3 0

3 years ago