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3 years ago

Find x- and y-intercepts. Write ordered pairs representing the points where the line crosses the axes.y=-2/5x+11/5

Mathematics

1 answer:

ivanzaharov [21]3 years ago

5 0

Answer:

y-intercept (0,11/5)

x-intercept (11/2,0)

Step-by-step explanation:

The y-intercept is obtained when the x-coordinate equals 0 (x=0)

so the y-intercept is (0,11/5)

The x-intercept is obtained when the y-coordinate equals 0 (y=0)

In this case

0 = -(2/5)x+11/5---->(2/5)x = 11/5 ----> x=(5*11)/(5*2) = 11/2

So the x-intercept is (11/2,0)

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Mashcka [7]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Helppp!!! Simplify this. Please explain it too. 18x^4 – 3x^3 – 2x^3 + x^2 – 10 + 6

strojnjashka [21]

Answer: 18x^4 - 5x^3 + x^2 - 4

Step-by-step explanation:

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3 years ago

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saveliy_v [14]

Answer:

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Step-by-step explanation:

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2 years ago

Find the cross product of <img src="https://tex.z-dn.net/?f=-%20%5Cfrac%7B3%7D%7B4%7Dv" id="TexFormula1" title="- \frac{3}{4}v"

dsp73

For any scalars $c_1,c_2$ , we have

$c_1\mathbf v\times c_2\mathbf w=c_1c_2\mathbf v\times\mathbf w$

So

$\left(-\dfrac34\mathbf v\right)\times\left(-\dfrac12\mathbf w\right)=\dfrac38\mathbf v\times\mathbf w$

We have

$\mathbf v\times\mathbf w=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-2&12&-3\\-7&4&-6\end{vmatrix}$
$=\begin{vmatrix}12&-3\\4&-6\end{vmatrix}\mathbf i-\begin{vmatrix}-2&-3\\-7&-6\end{vmatrix}\mathbf j+\begin{vmatrix}-2&12\\-7&4\end{vmatrix}\mathbf k$
$=-60\,\mathbf i-(-9)\,\mathbf j+76\,\mathbf k$
$=\begin{bmatrix}-60\\9\\76\end{bmatrix}$

which makes

$\left(-\dfrac34\mathbf v\right)\times\left(-\dfrac12\mathbf w\right)=\dfrac38\begin{bmatrix}-60\\9\\76\end{bmatrix}=\begin{bmatrix}-\frac{45}2\\\\\frac{27}8\\\\\frac{57}2\end{bmatrix}$

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3 years ago