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gavmur [86]
3 years ago
14

Find the percent of decrease from 280 to 210. Round to the nearest tenth of a percent, if necessary.

Mathematics
1 answer:
Vlada [557]3 years ago
5 0
To find the decreased percentage,here's what we can do:
\frac{originl \: value - new \: value}{originl \: value}  \times 100\%

In this case,the eqaution would be:
\frac{280 - 210}{280}  \times 100\% \\  =  \frac{70}{280}  \times 100\% \\  =  \frac{1}{4}  \times 100\% \\  = 25\%
Hope it helps!
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$289 is better because its cheaper per mile

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3 years ago
What is the simplified form of the following expression? 8ab/-4b
kondaur [170]

-2a

Step-by-step explanation:

steps:1. first write the question again : 8ab/-4b

2.first 8 divide by -4=-2

3.write ' a' with -2because u don't have another 'a' down belowso that u can pair it up and cancel

so: -2a.

4. cancel 'b' on top and 'b' at the bottom.

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3 years ago
PLS HELP I DONT REMEMBER HOW TO DO THIS
Gala2k [10]

Answer:

oh lord...ill try my best

Step-by-step explanation:

Let's start b writing down coordinates of all points:

A(0,0,0)

B(0,5,0)

C(3,5,0)

D(3,0,0)

E(3,0,4)

F(0,0,4)

G(0,5,4)

H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate  changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)

Reflecting

A(0,0,0)

B(0,5,0)

Reflecting

B(0,-5,0)

C(3,5,0)

Reflecting

C(3,-5,0)

D(3,0,0)

Reflecting

D(3,0,0)

b.)

A(0,0,0)

Moving

A(-2,-3,1)

B(0,-5,0)

Moving

B(-2,-8,1)

C(3,-5,0)

Moving

C(1,-8,1)

D(3,0,0)

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im sorry think this is wrong I TRIED MY BESTTTT!!!!!!!! give me credit this is tiring

6 0
3 years 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><u>Solution:</u></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

<em><u>The area of rectangle is given as:</u></em>

\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

<em><u>General form of quadratic equation is  </u></em>

{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.

3 0
3 years ago
50 POINTS
mamaluj [8]

Answer:

The answer is below

Step-by-step explanation:

The linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in seconds: A linear model with ordered pairs at 0, 60 and 2, 75 and 4, 75 and 6, 40 and 8, 20 and 10, 0 and 12, 0 and 14, 0. The x axis is labeled Time in seconds, and the y axis is labeled Height in feet. Part A: During what interval(s) of the domain is the water balloon's height increasing? (2 points) Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points) Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points) Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds.

Answer:

Part A:

Between 0 and 2 seconds, the height of the balloon increases from 60 feet to 75 feet  at a rate of 7.5 ft/s

Part B:

Between 2 and 4 seconds, the height stays constant at 75 feet.

Part C:

Between 4 and 6 seconds, the height of the balloon decreases from 75 feet to 40 feet at a rate of -17.5 ft/s

Between 6 and 8 seconds, the height of the balloon decreases from 40 feet to 20 feet at a rate of -10 ft/s

Between 8 and 10 seconds, the height of the balloon decreases from 20 feet to 0 feet at a rate of -10 ft/s

Hence it fastest decreasing rate is -17.5 ft/s which is between 4 to 6 seconds.

Part D:

From 10 seconds, the balloon is at the ground (0 feet), it continues to remain at 0 feet even at 16 seconds.

3 0
3 years ago
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