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3 years ago

9/3

Mathematics

1 answer:

hjlf3 years ago

8 0

Answer:

$960

Step-by-step explanation:

Given data

Principal=$800

rate= 4%

time= 5years

Applying the compound interest expression we have]

A=P(1+rt)

substitute

A=800(1+0.04*5)

A=800(1+0.2)

A=800*1.2

A=$960

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Quadrilateral A'B'C'D' is the image of quadrilateral ABC D under a dilation with a scale factor of 4. c What is the length of se

DENIUS [597]

a dilation with a scale factor of 4, is multiplication by 4 each side, so we need the measure of the side CD and then multiply by 4 to find C'D'

we can notice tht length of CD, is 3 squares,multiplying by 4 the new length is 3x4= 1 2 squares

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1 year ago

Hey guys, I need help with these questions. I need them done by today! Please, help :( There are 4 questions or more I think, I

Orlov [11]

Answer:

Step-by-step explanation:

Domain x^2 - 9 {Solution: - infinity < x < infinity}

Interval notation (- infinity, infinity)

Range of x^2 - 9 (Solution: f(x) is greater than or equal to - 9)

Interval notation (-9, infinity)

Axis interception points of x^2 - 9:

X- intercepts (3, 0) (-3, 0)

Y-intercepts (0, -9)

Vertex of x^2 - 9: Minimum (0, -9)

Solve for f:

f (x) = x^2 - 9

Step 1: Divide both sides by x.

fx / x = x^2 - 9 / x

f = x^2 - 9 / x

Answer:

f = x^2 - 9 / x

8 0

1 year ago

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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trasher [3.6K]

Answer:

the greatest common factor is 1.

Step-by-step explanation:

The factors of 3 are: 1, 3

The factors of 4 are: 1, 2, 4

The factors of 5 are: 1, 5

The factors of 17 are: 1, 17

Then the greatest common factor is 1.

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