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3 years ago

Please help me answer these questions I don’t know.

Mathematics

1 answer:

klemol [59]3 years ago

6 0

Answer:

y = -3x -3 (the graph is a straight line)

m = -3

b = -3

y = 2x + 2

m = 2

b = 2

2x - y = 6

x + y = 6

_______

3x = 12

3x/3 = 12/3

x = 4

Plug x into the equation to solve for y.

2(4) - y = 6

8 - y = 6

- y = 6 - 8

- y = -2

y = 2

Your solution is ( 4, 2).

3. No the solution is ( 10 , -11/2)

- 2x - 4y = 2

3x + 4y = 8

__________

x = 10

- 2 (10) - 4y = 2

- 20 - 4y = 2

- 4y = 22

- 4y/ 4 = 22/4

y = -11/2

Bonus:

4x + 3 ( 1787 - x) = 5792

4x + 5361 - 3x = 5792

x + 5361 = 5792

x = 431 (reserved seat tickets)

Minus 1787 with 431 which is 1,356 general admission tickets.

Answer: 431 reserved seat tickets and 1,356 general admission tickets!

Hope this helps!

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Solve the polynomial: 2x³-11x²+18x-8=0

kotykmax [81]

Answer:

2/0.74

Step-by-step explanation:

To solve the polynomial, factorize it.

2x³-11x²+18x-8=0

After factorizing:

(x-2)(2x^2 - 7x +4) = 0

now, solve each bracket one by one:

x-2 = 0

x = 2

Then

2x^2 - 7x + 4 = 0

use quadratic formula t get 0.74

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A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions

faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3>Solution:</h3>

Given that

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

The area of rectangle is given as:

$\text {Area of rectangle }=length \times width$

Area of rectangular parking lot = length of rectangular parking plot $\times$ width of the rectangular parking

$\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}$

But it is given that Area of rectangular parking lot = 15000 square feet

$\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}$

Solving the above quadratic equation using quadratic formula

General form of quadratic equation is

${ax^{2}+\mathrm{b} x+\mathrm{c}=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}$

$\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}$

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

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