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2 years ago

How do I solve 5-2(y+1)=21

1 answer:

8 0

5-2(y+1)=21

Distribute
5-2y-2= 21

Combine like terms
3-2y=21

Subtract 3 from both sides
-2y= 18

Divide both sides by -2
y= -9

Final answer: y=-9

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Which results only in a horizontal compression of y= 1/х by a factor of 6?

marta [7]

Answer:

I'm assuming you mean a compression of factor 6

In that case, it will be h = 1/6x

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3 years ago

Evaluate the expression for x = 3, y = 1/3 , and z = 5.<br><br> 12x − 3y / 4x − z

bixtya [17]

Answer:

30.25 in decimal form

30 1/4 in fraction form

Step-by-step explanation:

12(3) − 3(1/3) / 4(3)− (5) = 30.25 or 30 1/4

8 0

2 years ago

Read 2 more answers

Two friends, Hailey and Jenna bought two tickets to see a play at a local theater . Each ticket cost $17.25. Each friend donated

Sliva [168]

Per ticket cost $17.25

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3 years ago

Which one is prime number 2, 3, 5, 7, 13, 14, 15, 23, 25, 29, 30, 36, 61

soldier1979 [14.2K]

2,3,5,7,13,15,23,and 61 are all prime

5 0

3 years ago

MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) t

ioda

Answer:

We would have

$l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}$

where " l " is length, " w" is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

Surface area for a box with no top = $lw+2lh+2wh = 64$

where " l " is length, " w" is width and "h" is height.

Step 2.

Remember as well that

Volume of the box = $l*w*h$

Step 3

We can now use lagrange multipliers. Lets say,

$F(l,w,h) = lwh$

and

$g(l,w,h) = lw+2lh+2wh = 64$

By the lagrange multipliers method we know that

$\nabla F = \lambda \nabla g$

Step 4

Remember that

$\nabla F = (wh,lh,lw)$

and

$\nabla g = (w+2h,l+2h , 2w+2l)$

So basically you will have the system of equations

$wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)$

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

$lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)$

Then you would get

$l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)$

You can get rid of $\lambda$ from these equations and you would get

$lw+2lh = lw+2wh = 2wh+2lh$

And from those equations you would get

$l = w =2h$

Now remember the original equation

$lw+2lh+2wh = 64$

If we plug in what we just got, we would have

$l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}$

7 0

3 years ago