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2 years ago

Find an equation of a line with slope -3/4 and that passes through P(-4,6) in standard, form.

Mathematics

1 answer:

SashulF [63]2 years ago

3 0

Answer:

$y = - \frac{3}{4}x + 3$

Step-by-step explanation:

the standard form of any straight line is:

y = m × x + c

where: y is the y value of the point

x is the x value of the point

m is the gradient/slope

SOLUTION

$y = mx + c \\ 6 = - \frac{3}{4} ( - 4) + c \\ c = 3$

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Answer:

B. bisector of an angle.

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2 years ago

Reason abstractly. Are all horizontal lines parallel? Use slope<br> explain how you know.

ohaa [14]

Answer:

All horizontal lines are NOT parallel.

Step-by-step explanation:

We learned that lines go on forever.

If one line is slightly slanted, and the other line is completely straight, these lines are not parallel, as they will meet at some point.

Hope this helped :)

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3 years ago

How many solutions does this system have?

BigorU [14]

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3 years ago

Read 2 more answers

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2 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago