All categories

2 years ago

Please help me I really need help

Mathematics

1 answer:

Oliga [24]2 years ago

5 0

Answer:

Step-by-step explanation:

(1) 2x - 6y = -12

(2) x + 2y = 14

There is a -6y and a +2y. Since they have opposite signs, I'll try to eliminate the y terms. (That's my choice. There is more than one way to solve these.)

Multiply eq. (2) by 3:

3x + 6y = 42

Then add the result to eq. (1) to eliminate the y terms:

2x - 6y = -12

3x + 6y = 42

------------------

5x = 30, so x = 6

Now plug the value of x into eq. (2) and solve for y:

6 + 2y = 14

2y = 8

y = 4

Why did I use eq. (2) to solve for y? Because it's less work. I could have used eq. (1) instead:

2(6) - 6y = -12

12 - 6y = -12

-6y = -24

y = 4

More than one way to solve.

You might be interested in

A container of club soda has a volume of 3200 cubic meter. How many liters of soda does the container hold

svetoff [14.1K]

Answer:

3200000

Step-by-step explanation:

multiply the volume value by 1000

7 0

2 years ago

Proportions and equations solve

kirza4 [7]

Y=9x because if you divided y by x all the numbers will equal to 9 which is proportional

7 0

2 years ago

What's 49 divided by 261

Tpy6a [65]

.19
Hope this helped!

8 0

3 years ago

Read 2 more answers

Please Answer ASAP!

natali 33 [55]

Population. Population is the amount of whatever is inisde

4 0

3 years ago

Read 2 more answers

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago