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3 years ago

Graphing Absolute Value Equations

Mathematics

2 answers:

mr_godi [17]3 years ago

7 0

Answer:

The way I would like to solve this would be by using transformations. The standard form for transformation is a|b(x-h)|+k. Here, " - " is the a. Whenever a is < 0, it reflects across x-axis. The opposite sign of C is the value of vertical symmetry. So, vertical symmetry is, -(-4), which is, 4. (-(c), d ) is how you find the vertex. Here, opposite of c is 4. There is no d so it will be 0. So, the vertex is (4,0)

Step-by-step explanation:

Tip : Hey dudes and dudedas, use the following link to graph the any equations using online calculator. It will make your life, much more easier.

w-w-w desmos cam / practice

Thank me later.

Note : I disorted the link but I am sure that you are smart enough to find it out or type desmos in google.

MrMuchimi3 years ago

6 0

One way is to use transformations, so if you did then move the absolute value equation 4 units to the right then reflect it over the x axis

anyway
the vertex is at (4,0)
the line to the left of (4,0) is y=x-4
the line on the right is y=-x+4

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Answer:
3x - 12 is an expression equivalent to 3(x - 4)

Explanation:
Another way to write this expression is to simplify it. To simplify the given expression, multiply the number three by the x and negative four inside the parentheses.

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Therefore, 3x - 12 is the simplified version of the expression 3(x - 4). In other words, 3x - 12 is an expression equivalent to 3(x - 4).

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3 years ago

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Answer:

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3 years ago

Simplify 4+3(7+4*2) - 100

bazaltina [42]

Answer:

4+3(7+4*2) - 100 = -51

Step-by-step explanation:

4+3(7+4*2) - 100

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4 years ago

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4 years ago

A college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student i

kolbaska11 [484]

Answer:

Yes, Students are scoring above 2222 on the math portion of the exam.

Step-by-step explanation:

We are given that a college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student is ready for college-level mathematics i.e., population mean is 22.

Null Hypothesis, $H_0$ : $\mu$ = 22 {Students are scoring equal to 2222 on the math portion of the exam}

Alternate Hypothesis, $H_0$ : $\mu$ > 22 {Students are scoring above 2222 on the math portion of the exam}

Also, a random sample of 150 students who completed this core set of courses results in a mean math score of 22.7 on the college entrance exam with a standard deviation of 3.93 i.e.;

Sample mean, $Xbar$ = 22.7 and Sample standard deviation, s = 3.93

The Test statistics is given by;

Z = $\frac{Xbar -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

Test Statistics = $\frac{22.7 -22}{\frac{3.93}{\sqrt{150} } }$ ~ $t_1_4_9$

= 2.1815

Since, we are not given with the significance level, so we assume it to be 5%. At 5% significance level t table gives critical value of 1.6578 at 149 degree of freedom. Since our test statistics is more than the critical value so we reject null hypothesis as our test statistics fall in the rejection region.

Therefore, we conclude that Students are scoring above 2222 on the math portion of the exam.

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3 years ago