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goblinko [34]
2 years ago
12

The length of a rectangular poster is 9 more inches than two times its width. The area of the poster is 161 square inches. Solve

for the dimensions (length and width) of the poster.
Mathematics
2 answers:
irinina [24]2 years ago
8 0

Answer:

<u>Width = 7 inches</u>

<u>Length = 23 inches</u>

Step-by-step explanation:

Let :

  • length = 2x + 9
  • width = x

Equating to area :

  • (2x + 9)(x) = 161
  • 2x² + 9x = 161
  • 2x² + 9x - 161 = 0
  • 2x² - 14x + 23x - 161 = 0
  • 2x(x - 7) + 23(x - 7) = 0
  • x - 7 = 0
  • x = 7

Dimensions are :

  • <u>Width = 7 inches</u>
  • Length = 2(7) + 9
  • <u>Length = 23 inches</u>
BaLLatris [955]2 years ago
7 0

Answer:

Length: 23 inches

Width: 7 inches

Explanation:

  • Width: w
  • Length: 2w + 9
<h3>Area of rectangle: Length × Width</h3>
  • (2w + 9) × (w) = 161
  • 2w² + 9w - 161 = 0
  • 2w² + 23w - 14w -161 = 0
  • w(2w + 23) -7(2w +23) = 0
  • w = 7, -11.5

As <u>length measure cannot be negative</u>. Width is +7 inches

Length: 2w + 9 = 2(7) + 9 = 14 + 9 = 23

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Pie

Answer:

x - 2y = 2

Step-by-step explanation:

the equation of a line in standard form is

Ax + By = C ( A is a positive integer and A, B are integers )

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

multiply through by 2 to eliminate the fraction

2y = x - 2 ( subtract 2y from both sides )

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x - 2y = 2 ← in standard form


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vazorg [7]

Answer:

A, C ,E

Step-by-step explanation:

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2 years ago
A coin is tossed four times What is the probability of getting four heads? (1 mark) a) b) What is the probability of getting exa
BlackZzzverrR [31]
<h2>Answer:</h2>

1. A coin is tossed four times. What is the probability of getting four heads?

Each toss has a 1/2 chance of getting a head.

So the chance of getting all four heads can be calculated as :

1/2\times1/2\times1/2\times1/2=1/16

2. A coin is tossed four times. What is the probability of getting two heads?

Each toss can have 2 results, so 4 flips will have 2^{4} or 16 results.

Getting two heads means getting two tails also. So, we can get the number of times two heads come = \frac{4!}{2!2!} = 6

We can write the groups like - {HHTT,HTHT,HTTH,THHT,THTH,TTHH}

So, the required probability is : 6/16 or 3/8.

3. Not getting two heads means getting 3 tails and 1 head or all tails.

Probability of having all tails = 1/2\times1/2\times1/2\times1/2=1/16

Probability of one head(1st trial) and three tails = 1/16

Probability of one head (2nd trial) and three tails (the 1st, 3rd and 4th trials) = 1/16

Probability of one head (3rd trial) and three tails (the 1st, 2nd and 4th trials) = 1/16

Probability of one head (4th trial) and three tails (the 1st, 2nd and 3rd trials) = 1/16

So, the total probability showing only one or none head and at least three tails = 1/16+1/16+1/16+1/16+1/16=5/16

6 0
3 years ago
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