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Troyanec [42]
3 years ago
15

Over time, the value of the property at 397 West Lake Street increased by 275%. If the initial value of the property was P, whic

h expression represents the property’s current value?
Mathematics
1 answer:
VMariaS [17]3 years ago
6 0

Answer:

The expression is 2.75P.

Step-by-step explanation:

If P represents the initial value and the value increases by 275%, then you can say that it is worth 2.75 times as much as the initial value.  

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HELP awarding 60 points and brainliest answer!!!
Tju [1.3M]

Answer:

A

Step-by-step explanation:

Choice A is right because there is no dot above the two.

Choice B is not right because there is 2 dots above 1 and not 3.

Choice C is not right because the data doesn't go passed 5 to get to 8

Choice D is not right because it was 5 games because there is 5 dots above 3

5 0
3 years ago
Read 2 more answers
10times10times10times10times10times10
BARSIC [14]

Answer:

1000000

Step-by-step explanation:

10x10x10x10x10x10=1000000

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3 years ago
Which algebra expression represents this phase ?
lapo4ka [179]

Answer:

B

Step-by-step explanation:

Product is what you get when you multiply

5 0
3 years ago
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I got zero, but that’s wrong.
meriva

Answer:

Expected value is 4,000

Step-by-step explanation:

To find expected value ⇒ multiply the value by it's probability

40% × ( -25,000 ) = - 10,000

Breaking means neither add nor subtract a given amount

20% × 0 = 0

35% × 40,000 = 14,000

∴ Expected value = -10,000 + 14,000 = 4,000

LOSS:                   -10,000

BREAK EVEN:       0

WIN:                       14,000

Expected value is 4,000

<em>hope this helps....</em>

3 0
3 years ago
The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r
mash [69]

Answer:

Area = -2.3147

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates (2,0)

The area is represented as:

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Where

$r = 1 + \cos \theta$

and

(a,b) = (2,0)

Substitute values for r, a and b in

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta

Expand

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta

By integratin the above, we get:

Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]

Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]

Substitute 0 and 2 for \theta one after the other

Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}

Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  \frac{-sin(2)(cos(2) + 4) - 6}{4}

Get sin(2) and cos(2) in radians

Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}

Area = \frac{-9.2588}{4}

Area = -2.3147

3 0
2 years ago
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