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Svetllana [295]
3 years ago
14

20-90x-50x^2 find the vertex

Mathematics
2 answers:
Leni [432]3 years ago
6 0

20-90x-50x^2, written in standard form, would be -50x^2 - 90x + 20. Let's use the method of completing the square to determine the vertex:

Divide all terms in -50x^2 - 90x + 20 by -50, obtaining -50(x^2 + (9/5)x - 2/5)

The coefficient of x is 9/5. Divide this by 2, obtaining 9/10, and then square this result: (9/10)^2 = 81/100. Add this to x^2 + (9/5)x - 2/5 and then subtract it, resulting in x^2 + (9/5)x - 2/5 + 81/100 - 81/100, or

x^2 + (9/5)x + 81/100 - 2/5 - 81/100, or

(x + 9/10)^2 - 12100

Then the vertex location is (-9/10, -12100).

A faster approach would be to use x = -b / (2a) to determine the x-coordinate of the vertex and then to evaluate the function at this x-value to determine the y-coordinate. Here x = -(-90) / (-100) = -9/10 (same as before).

lawyer [7]3 years ago
4 0
I think it’s your answer

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4 years ago
If the odds against an event are 3:5, then the probability that the event will fail to occur is
damaskus [11]

Answer:

3/8

Step-by-step explanation:

probability = wanted outcomes / total outcomes

odds = wanted outcomes / unwanted outcomes

Odds of 3:5 losing means 3 losing outcomes and 5 winning outcomes.

The total outcomes is 8

The probability of losing which is the probability that the event will fail to occur is 3/8.

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2 years ago
Write an equivalent fraction for the whole number 6?
Keith_Richards [23]
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12/2 = 6

So, the answer could be 12/2
6 0
3 years ago
If the cost of a raffle ticket is $2, which of the following expressions could be used to represent the cost of m tickets? 2m 2
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6 0
3 years ago
A farmer has a 100 ft by 200 ft rectangular field that he wants to increase by 15.5% by cultivating a strip of uniform width aro
Advocard [28]

Answer:

(a) The strip should be 5ft wide

(b) The strip around the outside field is 10ft wide.

Step-by-step explanation:

Given:

Length of the rectangular field, L= 200 ft

width of the rectangular field, w = 100 ft

Area of the rectangular field, A = 200ft x 100ft = 20000 ft^2

let the width of the strip = x

The strip around the outside field = 2x

If the field is increased by 15.5%

New area of the field = 1.155 x 20000 = 23,100 ft^2

The increase in area of the field = 3,100 ft

3,100 = New area of field - old area of the field

3100 = (200 + 2x)(100 + 2x) - 20000

3100 = 20000 + 400x 200x + 4x^2 - 20000

3100 = 600x + 4x^2

Divide through by 4

775 = 150x + x^2

x^2 + 150x - 775 = 0

Factorize

(x + 155)(x-5) = 0

x = 5 ft

The strip should be 5ft wide.

The strip around the outside field = 2 x 5 ft = 10 ft

Thus, the strip around the outside field is 10ft wide.

3 0
3 years ago
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