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JulsSmile [24]
3 years ago
13

Write an absolute value inequality that represents the situation. The area, A, in square inches of a square photo is required to

satisfy 8.5 < A < 8.9.
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
7 0

Answer:

|A - 8.7| < 0.4

Step-by-step explanation:

Given

8.5 < A < 8.9

Required

Write as an absolute value inequality

The absolute value inequality can be derived using

|A - Mid| < Length

First Step: Get the middle of the interval

Mid = \frac{1}{2}(8.5 + 8.9)

Mid = \frac{1}{2}(17.4)

Mid = 8.7

Next: Get the Length of the interval

Length = 8.9 - 8.5

Length = 0.4

Substitute values for Mid and Length in  |A - Mid| < Length

|A - 8.7| < 0.4

Hence:

<em>The absolute inequality is </em>|A - 8.7| < 0.4<em></em>

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Answer:

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Step-by-step explanation:

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viktelen [127]
The answer is 1 1/15
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Jarrett's puppy weighed 3 3/4 ounces at Birth. At 1 week old, the puppy weighed 5 1/8 ounces. At 2 weeks old, the puppy weighed
Ad libitum [116K]

Answer:  The answer is 7\dfrac{7}{8} ounces.

Step-by-step explanation: Given in the question the weight of Jarrett's puppy at birth, one week old and 2 weeks old are as follows:

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We can see that

b_1-b_0=b_2-b_1=\dfrac{11}{8}.

So, the weight of the baby will make an arithmetic progression with first term 'a' and common difference 'd' as follows:

b_0=\dfrac{15}{4},~~d=\dfrac{11}{8}.

Thus, the weight of the puupy after 3 weeks will be

b_3=b_0+3\times d=\dfrac{15}{4}+3\times\dfrac{11}{8}=\dfrac{30+33}{8}=\dfrac{63}{8}=7\dfrac{7}{8}~\textup{ounces}.

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ddd [48]

The correct answer is D.

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Hope this helps.

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alexandr1967 [171]

Answer:

Step-by-step explanation:

11) First write in decreasing exponential terms and fill in blanks with zeros.

Our goal is to eliminate all term in the dividend by subtraction

          ________________

2v - 2 |  2v³ - 16v² + 0v + 13

we see that 2v needs to be multiplied by 1v² to eliminate the first term

          <u>              v²                </u>

2v - 2 |  2v³ - 16v² + 0v + 13

         <u>- (2v³ - 2v²)</u>

              0  - 14v²

multiply your estimate by your divisor and subtract from the dividend.

bring down the next term and repeat.

          <u>              v²    -7v       </u>

2v - 2 |  2v³ - 16v² + 0v + 13

         <u>- (2v³ - 2v²)</u>

              0  - 14v² + 0v

                 <u>-(-14v² + 14v)</u>

                             - 14v

repeat again

<u />

          <u>              v²  -  7v  -  7 </u>

2v - 2 |  2v³ - 16v² + 0v + 13

         <u>- (2v³ - 2v²)</u>

              0  - 14v² + 0v

                 <u>-(-14v² + 14v)</u>

                             - 14v + 13

                            <u>-(-14v + 14)</u>

                                          -1

and remainder gets put over the divisor and appended

v²  -  7v  -  7 - 1/(2v - 2)

13) Same process

          <u>                                      </u>

5a + 1 |  40a³ - 12a² - 39a - 5

          <u>              8a²                 </u>

5a + 1 |  40a³ - 12a² - 39a - 5

          <u>-(40a³ + 8a²)</u>

                     -20a² - 39a

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5a + 1 |  40a³ - 12a² - 39a - 5

          <u>-(40a³ + 8a²)</u>

                     -20a² - 39a

                   -(-<u>20a² -   4a)</u>

                                -35a - 5

          <u>              8a² -    4a  - 7  </u>

5a + 1 |  40a³ - 12a² - 39a - 5

          <u>-(40a³ + 8a²)</u>

                     -20a² - 39a

                    -(<u>20a² -   4a)</u>

                                -35a - 5

                              -<u>(-35a - 7)</u>

                                           2

8a² - 4a - 7 + 2/(5a + 1)

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2 years ago
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