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3 years ago

Find the ratio between 450ml and 1 litre

Mathematics

1 answer:

const2013 [10]3 years ago

5 0

Answer:

9:20

Step-by-step explanation:

Make both the given in same quantity i.e. ml and then you can reduce it to the lowest term and you get the answer to your question.

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The equation for the line g can be written as x-y=9. Line h is parallel to line g and passes through (4,5). What is the equation

Gnoma [55]

Answer:

y = x+1

Step-by-step explanation

The equation in slope intercept form is expressed as y = MX+c

m a the slope

Get the slope of the line

Given the line x-y =9

-y = -x+9

y = x-9

Slope of the line is 1

Since it is parallel to the given line, the slope o the required line is also 1

Substitute m = 1 adnd (4,5) into the point slope form

y-y0 =m(x-x0)

y - 5 = 1(x-4)

y-5 = x-4

y= x-4+5

y = x+1

Hence the required equation is y = x+1

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3 years ago

Find the solution to the following system using substitution. y = - 4 + 9 y = 3x - 5

lbvjy [14]

Answer:

1/2, 11/6

Step-by-step explanation:

y=-4+9y

9y-y=4

8y=4

y=1/2

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3x=5+1/2=11/2

x=11/6

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2 years ago

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skad [1K]

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Step-by-step explanation:

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4 years ago

Simplify 2x + 5x + 4x

Drupady [299]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 0

3 years ago

Find the ratio between 450ml and 1 litre​

Find the ratio between 450ml and 1 litre