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Debora [2.8K]
3 years ago
11

Joan wants to buy a rug for a room that is 1111 feet by 1414 feet. she wants to leave a uniform strip of floor around the rug. s

he can afford 8888 square feet of carpeting. what dimensions should the rug​ have?
Mathematics
1 answer:
pychu [463]3 years ago
4 0

Solution:

Let  x be the width of the strip.

Width of rug = 1111 - 2x

Length of rug = 1414 - 2x

Area of carpet = 8888

Area = Length * Width

8888 = (1414 - 2x) (1111 - 2x)

8888 = 1570954 - 2222x - 2828x + 4x²

4x² - 5050x + 1562066 = 0

2x² - 2525x + 781033 = 0

x = 720.5, 542

So, the value of x is 542.

Width of rug = 1111 - 2(542) = 27

Length of rug = 1414 - 2(542) = 330

Hence, the dimensions are 330, 27.

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torisob [31]
What is the missing sign between 3 and 2? I assume that it's multiplication sign, or *

3*2=6 (twice three is 6, because 3+3=6)

so the result of 3*2x is 6x, because 2x is actually 2*x, so we have 3*2*x - we are allowed to simplify this!

and 6 is the coefficient: the number by which a variable is multiplied.
3 0
3 years ago
From 12x ­- 6y + 9z subtract -x­ -3z + 6y . A)   12z ­- 13x -­ 12y B)   13x + 12y + 12z C)  - ­13x -­ 12y ­- 12z D)   13x -­ 12y
seraphim [82]
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4 0
3 years ago
Solving Quadratic Equations
NISA [10]

Answer:

a) x= 5 or x= -3

b) x = -4 + \sqrt{26} or -4 - \sqrt{26}

Step-by-step explanation:

a) x^{2} - 2x - 15 =0

x = \frac{-(-2) +\sqrt{(-2)^{2} - 4(1)(-15)}}{2(1)}  or  \frac{-(-2) -\sqrt{(-2)^{2} - 4(1)(-15)}}{2(1)}

x= 5 or x=-3

b) x^{2} + 8x - 10 =0

x = \frac{-(8) +\sqrt{(8)^{2} - 4(1)(-10)}}{2(1)}  or  \frac{-(8) -\sqrt{(8)^{2} - 4(1)(-10)}}{2(1)}

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7 0
3 years ago
The tables show the number of chin-ups done by students in two different gym classes. 2nd Period 3 4 12 14 7 7 8 4 8 3 11 10 9 8
Mekhanik [1.2K]

Answer:

On average, students in the 4th period did more chin-ups than students in the 2nd period

Step-by-step explanation:

Given the table values as

2nd Period     3  4 12 14 7  7  8  4  8  3  11  10  9   8 10 4  8 7  13 9

4th Period      10 3 14 14 16 15 7 12 7  10 12  8  10  9 6  11  9 1   2 5

Find the average chin-ups done by students in the gym classes

For 2nd Period gym class, the average will be;

=sum of chin-ups ÷ number of students

=sum of chin-ups in 2nd period gym class=

=3 + 4 +12+ 14+ 7 + 7 + 8+  4 + 8 + 3+  11+  10+  9 +  8 +10+ 4+  8+ 7 + 13+ 9\\\\=159

=number of students=20

Average=159÷20

=\frac{159}{20} =7.95\\\\\\=8

For 4th Period gym class

Sum of chin-ups=

=10+ 3+ 14+ 14 +16 +15+ 7 +12 +7+ 10+ 12 +8+ 10+ 9+ 6+ 11+ 9 +1 +2 +5\\\\\\=181

Number of students=20

Average number of chin-ups in the 4th period gym class

=\frac{181}{20} =9.05\\\\\\=9

Average number of chin-ups in 2nd Period Class = 8

Average number of chin-ups in 4th Period Class= 9

Conclusion; On average students in the 4th period gym class made more chin-ups (9) than those in 2nd period gym class (8).

3 0
4 years ago
Read 2 more answers
Given: AB / BC
Ahat [919]

1. given

2. perpendicular lines makes a right angle

5. definition of complementary angles

sorry bout the others. idk how to word it so hopefully what I gave you helps

8 0
3 years ago
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