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3 years ago

7

Number 36 . It's algebra 1 and I can't figure out what there trying to say

1 answer:

Ilya [14]3 years ago

8 0

Each touchdown is worth 6 points & after receiving the touchdown, the team will try making another touchdown & if they make it, it will only count as a 1 point from there. Let me try making an equation.

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Solve for x in the equation x^2+20+100=36

Nesterboy [21]

This is what I found in a calculator

3 0

3 years ago

3) Lane and Zack purchased a house for $245,000 in 2016. The house loses 4% of its value each year,

alina1380 [7]

Answer:

$138,345

Step-by-step explanation:

This is a compound decline problem, which will be solve by the compound formula:

$F=P(1-r)^t$

Where

F is the future value (value of house at 2030, 14 years from 2016)

P is the present value ($245,000)

r is the rate of decline, in decimal (r = 4% = 4/100 = 0.04)

t is the time in years (2016 to 2030 is 14 years, so t = 14)

We substitute the known values and find F:

$F=P(1-r)^t\\F=245,000(1-0.04)^{14}\\F=245,000(0.96)^{14}F=138,344.96$

Rounding it up, it will be worth around $138,345 at 2030

8 0

3 years ago

Bryan cut two peaches. he cut one into sixths and one into fourths.bryan ate 5sixths of the first peach and his brother ate 3fou

Oduvanchick [21]

Bryan : 5/6 = 20/24
Brother : 3/4 = 18/24
Answer Bryan ate more

4 0

3 years ago

Michel is headed to his aunt's house. For the first 2 hours he drives at 55 mph. For the next hour, he drives 70 mph. For the fi

polet [3.4K]

Answer:

the answer is 280

Step-by-step explanation:

5 0

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