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2 years ago

12

Round to the nearest ten thousand and then subtract

1 answer:

scZoUnD [109]2 years ago

5 0

39,151 rounds to 39,000 and 30,619 rounds to 31,000, so the answer is approximately 8,000.

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MATH SMARTIES I NEED HELP i will give brainliest CAN YOU TELL ME IF MY ANSWER IS CORRECT AND IF NOT CORRECT CAN YOU TELL ME THE

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Answer: 4

Step-by-step explanation: Since the total number of games played in the graph is 12, if you half that and get six, you half the total number of wins as well and get 4. :]

3 0

2 years ago

Read 2 more answers

Need help please, also about probabilities

liberstina [14]

Answer:

Step-by-step explanation:

Good evening ,

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2 years ago

Jackson’s mom limits the amount of time he is allowed to play video games. After Jackson plays for 9minutes, his mom tells him t

jeka94

Answer:

21 minutes

Step-by-step explanation:

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2 years ago

5. In the morning, Kevin's cat eats 1/2 of a can of cat food. In the afternoon, the cat eats 1/4 of a can of food. In the evenin

ELEN [110]

Answer:

2 cans

Step-by-step explanation:

1/2 + 1/4 + 4/5 = 1.55

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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