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3 years ago

I can't solve this equation pleas help!x^2+2x+8=0

Mathematics

1 answer:

tiny-mole [99]3 years ago

8 0

$x^2+2x+8=0\\
x^2+2x+1+7=0\\
(x+1)^2=-7\\
x+1=-\sqrt{-7} \vee x+1=\sqrt{-7}\\
x=-1-\sqrt7 i \vee x=-1+\sqrt7 i$

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Find a function rule for the line that passes through the origin (0,0) and the point (5, - 13)

inessss [21]

Answer:

The Function rule for line is: $\mathbf{y=-\frac{13}{5}x}$

Option A is correct option.

Step-by-step explanation:

Find a function rule for the line that passes through the origin (0,0) and the point (5, - 13)

The function rule is of form: $\mathbf{y=mx+b}$

where m is slope and b is y-intercept

Finding slope:

The formula used to find slope is: $Slope=\frac{y_2-y_1}{x_2-x_1}$

We have $x_1=0, y_1=0, x_2=5, y_2=-13$

Putting values and finding slope

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{-13-0}{5-0}\\Slope=\frac{-13}{5}\\Slope=-\frac{13}{5}$

Finding y-intercept

We will use point(0,0) to find y-intercept

$y=mx+b\\0=-\frac{13}{5}(0)+b\\0=0+b\\b=0$

So, y-intercept is 0

The Function rule for line having slope m=-13/5 and y-intercept b=0:

$y=mx+b\\y=-\frac{13}{5}x+0\\y=-\frac{13}{5}x$

So, The Function rule for line is: $\mathbf{y=-\frac{13}{5}x}$

Option A is correct option.

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