Find x and y<br>x=_____<br>y=____

How would I solve <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20x-%5Cfrac%7B5%7D%7B3%7D%20%3D-%5Cfrac%7B1%7D%7B2%7D%

sdas [7]

Multiply both sides of the equation by 12. Move the variables to the left-hand side and change its sign. Move the constant to the left-hand side and change its sign. Collect like terms. Add the numbers. Divide both sides of the equation by 12.

1/2 x - 5= - 1/2 x + 19/4
6x -20= - 6x + 57
6x + 6x= 57 + 20
12x= 57 + 20
12x= 77
ANSWER
x = 77/12
Alternative Form .
x = 6 5/12, x= 6.416

7 0

3 years ago

Read 2 more answers

Use the distributive property to solve the equation 28-(3x+4)=2(x+6)+x. Show your work.

Gre4nikov [31]

Answer:

x=2

Step-by-step explanation:

28−(3x+4)=2(x+6)+x

We need to use the distributive property for the right side.

28−3x−4=(2)(x)+(2)(6)+x)

28−3x−4=2x+2+x

−3x+24=3x+12

From here we need to subtract 3x from both side.

−3x+24−3x=3x+12−3x

6x+24=12

Transfer +24 on the right side.

6x=12−24

6x=−12

Finally, divide both sides by −6

6x/-6 = -12/-6

x=2

6 0

2 years ago

Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

So

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

So

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

So

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

So

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

8 0

3 years ago

1. If the stores sell flour for $0.75 per 1 KG how much would it cost for 180G

MAXImum [283]

1) 135 2) 119.40 if you still need the rest reply back to this answer

5 0

3 years ago

The product the lies between__and__. 3×3/4?

cupoosta [38]

Saying 3/4 * 3 is the same thing as saying 3/4 of 3. Imagine you have 3 pies, each cut into 4 slices. That would be 3 * 4 = 12 slices of pie, right?
So when there's 12 slices, that would be 4/4 of the pies, since all the pieces are there.
To find out how many pies we would have with 3/4 of the pie left, we just need to multiply the original equation.

3 * 3/4 = 2 1/4

The product of those two numbers lies between 2 and 3.
Hope that helped =)

7 0

3 years ago

Find x and yx=_____y=_____​

Find x and yx=_y=_