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3 years ago

Need help with mymathlab problem

Mathematics

2 answers:

ddd [48]3 years ago

8 0

I cant see a picture

Goshia [24]3 years ago

7 0

I can’t see your math lab problem, it’s not possible unless you could have shown me the footage based on the problem

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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:

$\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\ \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Answer: Third option

8 0

1 year ago

What is the equation of the line that passes through the point (−3,2) and has an undefined slope?

kirza4 [7]

The answer is x=-3

Sure hope this helps you

4 0

2 years ago

A river has a current of 2km per hour. Find the rate of Fred’s boat in still water if it travels 30 km downstream and the same t

ratelena [41]

Answer:

32km

Step-by-step explanation:

8 0

3 years ago

Assume that the probability of a driver getting into an accident is 7.6%, the average cost of an accident is $16,412.05, and the

arlik [135]

Answer:

1,352.32

Step-by-step explanation:

If you take 7.6 and multiply it to 16,412.05 (.076)(16,412.05), you get 1,247.32. You then add the overhead cost of 105 to get 1352.32. (Apex verified too)

5 0

3 years ago

Read 2 more answers

Rectangle ABCD was dilated to create rectangle A'B'C'D.

77julia77 [94]

The first thing we must do for this case is to calculate the scale factor.

For this, we make the relationship between two parallel sides.

We have then:

$k =\frac{B'C'}{BC}$