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White raven [17]
3 years ago
8

A parking lot has 916 parking spaces that are shaped like parallelograms. The base of each space is 10 feet, and the height of e

ach space is 18.5 feet. What is the total area of the parking spaces?
A 185 ft2
B 944.5 ft2
C 26,016 ft2
D 169,460 ft2
Mathematics
2 answers:
VladimirAG [237]3 years ago
5 0
First find the area of each parking space
10*18.5
185 feet2
then multiply the area of each parking space by the total amount go parking spaces
185*916
169,460 ft2 is the total area of the parking spaces
MAVERICK [17]3 years ago
5 0

Answer:

Option D. 169460 ft²

Step-by-step explanation:

There are 916 parking spaces shaped in a parallelogram. Each space is 10 feet long and 18.5 feet high.

we have to calculate the total area of the parking spaces.

Since spaces are in the shape of parallelogram therefore formula to measure the area will be = Base × height

Area of one space = 10 × 18.5 = 185 ft²

There are 916 parking spaces so total area will be = 185×916 = 169460 ft²

Answer is Option D. 169460 square feet.

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The equation of the quadratic function is f(x) = x²+ 2/3x - 1/9

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Substitute the known values in the above equation

So, we have the following equation

f(x) = (x - \frac{-1-\sqrt{2}}{3})(x - \frac{-1+\sqrt{2}}{3})

This gives

f(x) = (x + \frac{1+\sqrt{2}}{3})(x + \frac{1-\sqrt{2}}{3})

Evaluate the products

f(x) = (x^2 + \frac{1+\sqrt{2}}{3}x + \frac{1-\sqrt{2}}{3}x + (\frac{1-\sqrt{2}}{3})(\frac{1+\sqrt{2}}{3})

Evaluate the like terms

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7 0
1 year ago
Please help me!<br> Find the number x such that f(x) =1
STatiana [176]

Answer:

D

Step-by-step explanation:

We have the piecewise function:

f(x) = \left\{        \begin{array}{ll}            -\frac{1}{2}x-1 & \quad x \leq -2 \\            x & \quad x > -2        \end{array}    \right.

And we want to find x such that f(x)=1.

So, let's substitute 1 for f(x):

1 = \left\{        \begin{array}{ll}            -\frac{1}{2}x-1 & \quad x \leq -2 \\            x & \quad x > -2        \end{array}    \right.

This has two equations. So, we can separate them into two separate cases. Namely:

1=-\frac{1}{2}x-1\text{ or } 1=x

Let's solve for x in each case.

Case I:

We have:

1=-\frac{1}{2}x-1

Add 1 to both sides:

2=-\frac{1}{2}x

Let's cancel out the fraction by multiplying both sides by -2. So:

2(-2)=(-2)\frac{-1}{2}x

The right side cancels:

-4=x\\

Flip:

x=-4

So, x is -4.

Case II:

We have:

1=x

Flip:

x=1

This is the solution for our second case.

So, we have:

x_1=-4\text{ or } x_2=1

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 <em>is </em>less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 <em>is</em> greater than or equal to -2. So, x=1 is <em>also </em>a solution.

Therefore, our two solutions are:

x_1=-4\text{ or } x_2=1

Out of our answer choices, we can pick D.

And we're done!

7 0
3 years ago
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