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3 years ago

7

Fraction 4 over 5 n = Fraction 2 over 3. n = ___?

1 answer:

Taya2010 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

(4/5)n = (2/3)

move 4/5 to the right

n = (2/3)/(4/5)

n = 5/6

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Please help with this algebra problem!! thanks

saw5 [17]

5.7t < 420 divide both sides by 5.7

5.7 5.7

t < 73.68

5 0

3 years ago

Read 2 more answers

How would I solve for x?

vaieri [72.5K]

Answer:

$x = 65$

Step-by-step explanation:

First, we calculate the unknown angle of the left triangle and right triangle: $180 - 62 - 50 = 68$

Unknown angle for left triangle⬆️

$180 - 80 - 53 = 47$

Unknown angle for right triangle⬆️

Now we've calculated the two angles for the small triangle in the middle. So, we just have to calculate the top angle of the middle triangle:

$180 - 68 - 47 = 65$

The angle 65° and angle of x are vertical angle.(both are equal)

So... <u>x= 65°</u>

<em>hope it </em><em>helps </em><em>:</em><em>)</em>

4 0

2 years ago

Furniture store is having a sale where everything is 40% off.Write a function that represents the amount of discount d on an ite

Vladimir79 [104]

Answer:

d = .4p

Step-by-step explanation:

Discount = original price * discount rate

We know the discount rate is 40%

original price is p

discount = d

d = p*40%

Changing this to decimal form

d = .4p

6 0

3 years ago

Find the slope of a<br> line with ordered pairs<br> (1,5) and (5,8)

klemol [59]

Step-by-step explanation:

(1,5) (5,8)

x,y. x,y

Formula: <u>y2-y1</u>

x2-x1

so...

<u>8-5</u><u>=</u> <u>3</u>

5-1 = 4

so... <u>3</u>

4

6 0

2 years ago

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

2 years ago