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2 years ago

Nancy planted 44 plants in rows. If there are 7 plants in each row, how many plants are left over

Mathematics

1 answer:

deff fn [24]2 years ago

8 0

Answer:There will be 2 plants left over.

Step-by-step explanation: 44 divided by 7 equal 6.2. The whole number is 6 and the 2 is the left over.

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Eight figures are below. Which ones are convex polygons?

Nataly_w [17]

Answer:

A, C, D, F, G. I believe so?

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2 years ago

Your neighbor has been saving to buy a dirt bike. He has found a dirt bike for $836, which is 88% of the total amount he has sav

Gre4nikov [31]

Answer:

$936.32

Step-by-step explanation:

To solve this problem we need to find the remaining 12% of money our neighbor has saved since 88% was $836. To do this, we simply multiply $836 by 12%. Remember to change 12% into its decimal form (12% -> 0.12).

$836 x 0.12 = $100.32

Now, our last step is to add $100.32 to $836.

$836 + $100.32 = $936.32

Our answer is $936.32.

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2 years ago

25. If the number 39 is included in the following data 19, 21, 25, 36, 30, 19,

lozanna [386]

It should be range your welcome

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2 years ago

Read 2 more answers

Khan khan khan khan

Ber [7]

Answer:

Step-by-step explanation:

it says motor cycles combined with garbage so add both of them then subtract that from the bull dosers

motor =18

garbage=4

18+4=22

bulldosers=11

22-11= 11 more

5 0

3 years ago

Read 2 more answers

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box $V=x^2z$

Volume of the box $=12 ft^3\\$

$Therefore, x^2z=12\\z=\frac{12}{x^2}$

The material for the base costs $\$0.17/ft^2$ , the material for the sides costs $\$0.10/ft^2$ , and the material for the top costs $\$0.13/ft^2$ .

Area of the base $=x^2$

Cost of the Base $=\$0.17x^2$

Area of the sides $=4xz$

Cost of the sides= $=\$0.10(4xz)$

Area of the Top $=x^2$

Cost of the Base $=\$0.13x^2$

Total Cost, $C(x,z) =0.17x^2+0.13x^2+0.10(4xz)$

Substituting $z=\frac{12}{x^2}$

$C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}$

To minimize C(x), we solve for the derivative and obtain its critical point

$C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2$

Recall: $z=\frac{12}{x^2}=\frac{12}{2^2}=3\\$

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

7 0

2 years ago