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kipiarov [429]
3 years ago
10

HHHEEEELLLPPPPP just 3 questions!!!!!!!!!!!!!!!!!!!!! 1. How does evaluating a function at specific points relate to graphing th

e function? 2. What is the process for finding x and y - intercepts from a graph? a table? an equation? 3. Can you think of a career or field of work that utilizes the skills of interpreting, comparing, and/or graphing functions? How might that career or field use the skill/concept?
Mathematics
1 answer:
Tema [17]3 years ago
4 0

Answer:

1) When we have a function y = f(x), evaluating this function in one given point (x0 for example ) is equivalent to changing the x inside the function by the value x0.

This will give us the point (x, f(x0)).

now, a graph is conformed by the set of all the points (x, f(x)).

2) The x and y-intercepts are the points where the graph intercepts the x-axis or the y-axis.

If you have a table with points (x, y), the y-intercept is the point (0, y') and the x-intercept is the point (x', 0)

Where x' and y' are known values.

If you have an equation:

y = f(x).

The x-intercept is the point (x', 0)

So you need to find the value x' such that f(x') = 0.

The y-intercept is the point (0, y')

So you need to evaluate the function in x = 0, and find the value of y.

y' = f(0).

3) The obvious answer is any science-related career, where graphs are used to transmit a lot of information with almost no words.

Another option can be an economist, where, for example, interpreting and comparing graphs can be useful to determine which option fits better the needs of a given company or what decision is better to be taken.

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A manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly.
denpristay [2]

Answer:

We conclude that the population mean light bulb life is at least 500 hours at the significance level of 0.01.

Step-by-step explanation:

We are given that a manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly. The population standard deviation is 50 hours and the light bulb life is normally distributed.

You select a sample of 100 light bulbs and find mean bulb life is 490 hours.

Let \mu = <u><em>population mean light bulb life.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 500 hours      {means that the population mean light bulb life is at least 500 hours}

Alternate Hypothesis, H_A : \mu < 500 hours     {means that the population mean light bulb life is below 500 hours}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                           T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean bulb life = 490 hours

           σ = population standard deviation = 50 hours

           n = sample of light bulbs = 100

So, <u><em>the test statistics</em></u>  =  \frac{490-500}{\frac{50}{\sqrt{100} } }

                                     =  -2

The value of z test statistics is -2.

<u>Now, at 0.01 significance level the z table gives critical value of -2.33 for left-tailed test.</u>

Since our test statistic is higher than the critical value of z as -2 > -2.33, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis.</u>

Therefore, we conclude that the population mean light bulb life is at least 500 hours.

7 0
3 years ago
What is the vertex of y = x2 + 2x +2<br> O (1,-1)<br> O (2,-2)<br> O (1, 2)<br> 0 (-1, 1)
fredd [130]

Answer:

vertex = (- 1, 1 )

Step-by-step explanation:

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

x = - \frac{b}{2a}

y = x² + 2x + 2 ← is in standard form

with a = 1 and b = 2 , then

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

substitute x = - 1 into the equation for y- coordinate of vertex

y = (- 1)² + 2(- 1) + 2 = 1 - 2 + 2 = 1

vertex = (- 1, 1 )

8 0
3 years ago
If (x, y) = (3, 4) is a solution of the simultaneous equations.
Vinvika [58]
The first thing you want to do is plug in x and y into both equations:

a(3) + b(4) = 4

b(3) + a(4) = 8

rearrange to line up a’s and b’s

3a + 4b = 4

4a + 3b = 8

now you want to choose a or b and multiply each equation by a number to make them have the same amount of a’s or b’s.

4(3a + 4b = 4) = 12a + 16b = 16

3(4a + 3b = 8) = 12a + 9b = 24


Now we subtract the bottom equation from the top and solve for b:

12a + 16b - (12a + 9b) = 16 - 24

7b = -8

b = -8/7


Now we plug back in for b to one of the original equations:

3a + 4(-8/7) = 4

3a + (-32/7) = 4

3a - (32/7) = 4

3a = 4 + (32/7)

3a = (28/7) + (32/7)

3a = 60/7

a = (60/7)/3 = 20/7.


Finally, plug a and b in together to double check using the second equation.

4a + 3b = 8

4(20/7) + 3(-8/7) = ?

(80/7) - (24/7) = ?

56/7 = 8.
4 0
3 years ago
In a geometric sequence, if a3 = ‒5 and a6 = 40, determine a1, r, and an. Then write the first three terms of the sequence.
Tresset [83]
a_4=ra_3
a_5=ra_4=r^2a_3
a_6=ra_5=r^3a_3

40=-5r^3\implies r^3=-8\implies r=-2

a_3=ra_2=r^2a_1

-5=(-2)^2a_1\implies a_1=-\dfrac54

You know the first and third terms, a_1 and a_3, so you just need to find the second term a_2:

a_2=ra_1\implies a_2=-2\left(-\dfrac54\right)=\dfrac52
7 0
3 years ago
Please solve the problem ​
jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.

Then you can eliminate <em>Y</em> by taking the sum

4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>

==>   <em>X</em> = (4<em>A</em> + <em>B</em>)/11

Similarly, you can eliminate <em>X</em> by using

-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>

==>   <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11

It follows that

X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}

Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

3 0
3 years ago
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