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Nana76 [90]
3 years ago
6

What is equivalent to (y+4)+10​

Mathematics
1 answer:
WITCHER [35]3 years ago
8 0

Answer:

brb

Step-by-step explanation:

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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-3x=\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
I hate people who answer questions like this:
fiasKO [112]

Answer:

Oh wow

Step-by-step explanation:

Nice opinion xD

6 0
3 years ago
Solve: -2/3a - 6 = 12 <br>a =​
Anika [276]

Answer: -27 is the correct answer. (apparently)

7 0
3 years ago
A portable electronic typewriter cost £180 in 1988. Its price rose by 5% in 1989. What was the new price?
elena-s [515]

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{5\% of 180}}{\left( \cfrac{5}{100} \right)180}\implies 9~\hfill \stackrel{\textit{new price}}{180+9\implies 189}

5 0
3 years ago
Write the equation in y=mx+b form please????
Solnce55 [7]

ANSWER

y =  - 2x - 5

EXPLANATION

We can use any two points to determine the slope.

The line passes through;

(-3,1) and (-5,5).

The slope is given by

m =  \frac{y_2-y_1}{x_2-x_1}

This implies that;

m =  \frac{5 - 1}{ - 5 -  - 3}

m =  \frac{4}{ - 2}  =  - 2

We can now find the equation using the formula;

y-y_1=m(x-x_1)

y   - 1 =  - 2(x  + 3)

Expand:

y =  - 2x -6 + 1

y =  - 2x - 5

6 0
3 years ago
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