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2 years ago

HELP PLS HELP PLS

Mathematics

1 answer:

ivanzaharov [21]2 years ago

5 0

Answer:

x =-4

y=-1

x=2

y=8

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X+2y=8<br>2x-y=1<br>mathe question

Ira Lisetskai [31]

Answer:

x = 2 and y = 3

Step-by-step explanation:

We have

x+2y = 8 -----equation (i)

2x-y = 1 -----equation (ii)

Now,

x+2y = 8

or, x = 8-2y ------equation (iii)

Again,

2x-y = 1

or, 2x = 1+y

or, 2x-1 = y

Putting the value of 'x' from equation (iii)

or, 2(8-2y)-1 = y

or, 16-4y-1 = y

or, 16-1 = y+4y

or, 15 = 5y

or, y = 15/5

or, y = 3

Putting the value of 'y' in equation (iii)

x = 8-2y

or, x = 8-(2×3)

or, x = 8-6

or, x = 2

Therefore, x = 2 and y = 3.

4 0

2 years ago

Follow the process of completing the square to solve 2x2 + 8x - 12 = 0. After adding B2 to both sides of the equation in step 4,

Katena32 [7]

2x^2 + 8x - 12 = 0..divide by 2
x^2 + 4x - 6 = 0
x^2 + 4x = 6...add 4 to both sides of the equation
x^2 + 4x + 4 = 6 + 4
(x + 2)^2 = 10....<== ur constant is 10
x + 2 = (+-)sqrt 10
x = -2 (+ - ) sqrt 10

x = -2 + sqrt 10
x = -2 - sqrt 10

3 0

3 years ago

Read 2 more answers

PLZ HELPPPP !!!

Blizzard [7]

Answer:

Step-by-step explanation:

Calculate the volume of a cone by its base and height with the equation volume = 1/3 * base * height. You can calculate the height of a cone from its volume by reversing this equation. Triple the volume amount. For this example, the volume is 100.

7 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago

The state highway patrol wants to evaluate a new rule for vehicles that was put into place to prevent road accidents across all

Mnenie [13.5K]

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3 0

3 years ago