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Leto [7]
3 years ago
7

The graph of an absolute value function contains the points (1, 2), (0, 0), and (–1, 2). Shawn thinks the function this graph re

presents is f(x) = 2|–x|. Brielle thinks the function this graph represents is f(x) = 2|x|. Determine who is correct and explain why.
Mathematics
2 answers:
KATRIN_1 [288]3 years ago
5 0

<span>In mathematics, the </span>absolute value<span> <span>or modulus. | x. | of a real number x is the non-negative </span></span>value<span> of x without regard to its sign. So both shawn and brielle are correct since both function will have the same answer, because it will just cancel out the sign.</span>

nekit [7.7K]3 years ago
3 0

Answer:

The absolute value parent function is symmetric with respect to the y-axis.  

The graphs of f(x) = 2|–x| and f(x) = 2|x| are identical because the absolute value of negative x is a reflection across the y-axis.  

If you substitute the three given points into both functions, they produce true statements.

Step-by-step explanation:

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Piece function:
Nimfa-mama [501]

First, make the table so you can sketch the graph:

    x   |    y

    5  |   20        

    4  |    9              x² - 5   ; x > 3

 <u>   3  |    </u><u>4 </u><u>  </u><u><em>(open dot)                      </em></u><em>   ⇒ </em>from the right

  <u>  3  |    </u><u>5</u><u>                  5     ; x = 3       </u>   ⇒ at x = 3

    3  |    2  <em>(open dot)                          ⇒ </em>from the left

    2  |    4              -2x + 8  ; x < 3          

<u>     1   |    6                                           </u>


Next, look at the graph (or table) to find the limits:

lim 3⁺ = 4  <em>as x approaches 3 from the right, y approaches 4</em>

lim 3⁻ = 2  <em>as x approaches 3 from the left, y approaches 2</em>

lim 3 = DNE <em>lim 3⁺ ≠ lim 3⁻ so the limit does not exist</em>

f(3) = 5  <em>when x = 3, y = 5</em>

f(x) is NOT continuous at x = 3 <em>because lim 3 ≠ f(3)</em>

5 0
3 years ago
I will give you BRAINLIEST for the correct answer
Alekssandra [29.7K]
6 x 18 x 20 = 2,160 divided by 3 equals 720
8 0
3 years ago
Perform the computation and write the result in scientific notation:
gtnhenbr [62]

Answer:

6.06 x 10^{9}

Step-by-step explanation:

(5.25 x 10^{5} ) / (8.7 x 10^{-5} )

= (5.25 x 10^{5} ) x( 10^{5} )/ (8.7 )

= (5.25 x 10^{5 + 5} )/ (8.7 )

= 0.606 x 10^{10}

= 6.06 x 10^{9}

6 0
3 years ago
What is the nth term of quadratic sequence 4 7 12 19 28
Anni [7]
We find the first differences between terms:
7-4=3; 12-7=5; 19-12=7; 28-19=9.
Since these are different, this is not linear.
We now find the second differences:
5-3=2; 7-5=2; 9-7=2. Then:

Since these are the same, this sequence is quadratic.
We use (1/2a)n², where a is the second difference:
(1/2*2)n²=1n².

We now use the term number of each term for n:
4 is the 1st term; 1*1²=1.
7 is the 2nd term; 1*2²=4.
12 is the 3rd term; 1*3²=9.
19 is the 4th term; 1*4²=16.
28 is the 5th term: 1*5²=25.

Now we find the difference between the actual terms of the sequence and the numbers we just found:

4-1=3; 7-4=3; 12-9=3; 19-16=3; 28-25=3.

Since this is constant, the sequence is in the form (1/2a)n²+d;
in our case, 1n²+d, and since d=3, 1n²+3.
The correct answer is n²+3
7 0
3 years ago
The table below shows data from a survey about the amount of time students spend doing homework each week. The students were in
nadezda [96]

Answer:

The correct option is;

Both spreads are best described by the standard deviation

Step-by-step explanation:

The given information are;

,                                    College                       High School

High,                              20                               20

Low,                                6                                 3

Q₁,                                   8                                 5.5

Q₃,                                  18                                16

IQR,                                 10                                10.5

Median,                           14                                11

Mean,                              13.3                             11

σ,                                      5.2                             5.4

Checking for outliers, we have

College

Q₁ - 1.5×IQR gives 8 - 1.5×10 = -7

Q₃ + 1.5×IQR gives 18 + 1.5×10 = 33

For high school

Q₁ - 1.5×IQR gives 5.5 - 1.5×10.5 = -10.25

Q₃ + 1.5×IQR gives 16 + 1.5×10.5 = 31.75

Therefore, there are no outliers and the data is representative of the population

From the data, for the college students, it is observed that the difference between the mean, 13.3 and Q₁, 8, and between Q₃, 18 and the mean,13.3 is approximately the standard deviation, σ, 5.2

The difference between the low and the high is also approximately 3 standard deviations

Therefore the college spread is best described by the standard deviation

Similarly for the high school students, the IQR is approximately two standard deviations, the  difference between the mean, 11 and Q₁, 5.5, and between Q₃, 16 and the mean,11 is approximately the standard deviation, σ, 5.4

Therefore the high school spread is also best described by the standard deviation.

4 0
3 years ago
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