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2 years ago

Expressions that have the same value

1 answer:

6 0

If you need an equivalent expression just make a simple problem like 30+4=4+30 then add all the complicated stuff needed to complete the problem

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What is the solution to this inequality?

Snezhnost [94]

Answer:

$\boxed{\sf{D. \quad x < 3}}$

Step-by-step explanation:

It is necessary to isolate the term of x from one side of an inequality to find its solution.

First, add by 8 from both sides.

$\sf{-4x-8+8 > -20+8}$

Solve.

-20+8=-12

Rewrite the problem down.

$\sf{-4x > -12}$

Multiply by -1 from both sides.

(-4x)(-1)<(-12)(-1)

Solve.

Multiply the numbers from left to right.

(-12)(-1)=12

4x<12

Divide by 4 from both sides.

4x/4<12/4

Solve.

Divide the numbers from left to right.

12/4=3

x<3

Therefore, the correct answer is D. x<3 (last option).

I hope this helps you! Let me know if my answer is wrong or not.

5 0

2 years ago

Read 2 more answers

I need a lot of help! Is 54.03 + 4.04 greater than 54 + 4.4?

Aliun [14]

Answer:

54+4.4 is greater

Step-by-step explanation:

if u need explanation than please ask

7 0

1 year ago

How many solutions does the equation 3x-2(x+10)=x-20

Ber [7]

Answer:

infinite

Step-by-step explanation:

5 0

2 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

If the area of traingle TQR is 60 cm square,what is the area of the parallelogram PQRS?

Stolb23 [73]

Answer:

pls mark me brainlist .......my question was the same but no one ans I don't know

8 0

1 year ago