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2 years ago

8

A contractor estimates $2,400 to build a deck to a customer’s specifications, How much would it cost to build five similar decks

?

1 answer:

Bumek [7]2 years ago

3 0

5*2400=$12,000 is the total costs for 5 decks costing $2,400 each.

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Graph y=2/5x, use x as -2,-1,0,1,2

Natali [406]

Plot a graph of y-axis against x-axis.

The graph should be a straight line passing through:

(-2,0.8) , (-1,-0.4) , (0,0) , (1,0.4) and (2,0.8)

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2 years ago

Point H is the midpoint of side FK. Triangles F G H and K J H are connected at point H. The lengths of sides F H and H K are con

earnstyle [38]

Answer:

the answer would be x=6

to find it you would set 3x-2 equal to 2x+4 and solve because the two sides would be equal if the triangles are congruent

hope this helps :)

Step-by-step explanation:

3 0

1 year ago

Read 2 more answers

12 is 75% of what number?

wolverine [178]

For this question you should say:
75/100 = 12/?
so the ? is your answer:
12*100/75 = 16 :)))
I hope this is helpful
have a nice day

8 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

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

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2 years ago

Balls colliding on a pool table are an example of how energy changes when objects collide. When two balls collide, the first bal

Lena [83]

5/10= 50/100 + 33/100= 83/100..... Does that help?

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2 years ago