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2 years ago

6

Create a system of linear equations that when solved algebraically, prove to have no solution.

1 answer:

ZanzabumX [31]2 years ago

4 0

Answer: y = 3x + 4 and y = 3x + 7

Step-by-step explanation:

You might be interested in

Twice a number is 12 more than five times the number

inna [77]

N x 2 = 12 + 5n
2n - 5n = 12
-3n = 12
3n = -12
n = -4

4 0

3 years ago

You are taking a quiz that has 9 multiple-choice questions. If each question has 4 possible answers, how many different ways are

11111nata11111 [884]

Answer:

9 questions x 4 answers per question

9x4= 36

6 0

3 years ago

Read 2 more answers

What is the volume of a box with a height of 3/2 inches, a length of 7/2 inches, and a width of 5/2 inches.

JulsSmile [24]

Answer:

13.125 cubic inches

Step-by-step explanation:

Volume of box is 3/2 X 7/2 X 5/2 = 105/8 = 13 1/8 = 13.125 cubic inches

4 0

3 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

Please help asap What is the sum of the following two polynomials?<br> (3x2 – 5x + 3) and (-x2 – 3x)

Anton [14]

(3x^2 - 5x +3) + (-2x -3x)
=3x^2-5x+3 + (6x^2)
=9x^2-5x+3

4 0

2 years ago

Read 2 more answers