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3 years ago

How do I do algebra

1 answer:

tatiyna3 years ago

5 0

Know the order of operations. One of the trickiest things about solving an algebra equation as a beginner is knowing where to start. Luckily, there's a specific order for solving these problems: first do any math operations in parentheses, then do exponents, then multiply, then divide, then add, and finally subtract. (Wikihow)
PEMDAS
parenthesis
exponents
multiply
divide
add
subtract

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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

Megan wants to retile her bathroom floor. If the home improvment store sel

Pavel [41]

79.68 dollars...........

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3 years ago

*Spam answers will not be tolerated*

V125BC [204]

Step-by-step explanation:

C = -11

A = -13

V = 9

E = -9

If you are adding them: -24

Hope this helps!

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2 years ago

Can someone help me with this? Please-

lozanna [386]

Answer:

Step-by-step explanation:

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2 years ago

Which of the following could be a rational number?

Flura [38]

Answer:

the sum of two irrational numbers

Step-by-step explanation:

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3 years ago