Someone please help me

Mathematics

1 answer:

lana66690 [7]4 years ago

4 0

I would help, nut the image doesn't seem to be clear.
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Acellus help slope intercept

Serjik [45]

Greetings.

The answer is y = 2x + 7.

Explanation:

If you are an expert in graph then you are able to find the slope without calculating. (Rise over Run by looking at the graph.)

We can also find the slope by calculating m-value.

$m=\frac{y_2-y_1}{x_2-x_1}$ ---> Rise Over Run

<em>But how are we gonna calculate if we don't have any two given ordered pairs?</em>

The answer is, we can just pick two ordered pairs that are part of the graph, or rather say that the graph passes through.

The graph passes through (-3,1) and (-2,3).

Then we can calculate the slope by using rise/run.

$m=\frac{3-1}{-2-(-3)}\\m=\frac{2}{-2+3}\\m=\frac{2}{1}\\m=2$

Thus, the slope is m = 2.

However, we are not done yet. We just only found out the slope. We need to find the y-intercept too.

From Linear Function - $y=mx+b$

Where m = slope and b = y-intercept.

Since b = y-intercept, we don't need to calculate/substitute x = 0. We simply find which number the graph has y-intercept at.

Which is (0,7), the y-intercept.

Then we substitute b = 7 as y-intercept.

Thus the answer is $y=2x+7$

3 0

3 years ago

I need you to answer with a, b, c, d

solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

$\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

For the given function:

$f(x)=2x^2-10x-3$ $x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}$ $x=\frac{10\pm\sqrt[]{100+24}}{4}$ $\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}$ $\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}$ $\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Then, the zeros of the given quadratic function are: