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3 years ago

PLS HELP how many solutions does this system of equations below have? y=8x+4 y=8x+3

Mathematics

1 answer:

serious [3.7K]3 years ago

5 0

There are to be exact only 2 ways to answer this equation because of the way math was made.

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Rewrite the following in the form log(c) log(6)−log(3)

Genrish500 [490]

<h3>Answer: log(2)</h3>

Explanation:

The rule to use here is log(x)-log(y) = log(x/y)

So,

log(6)-log(3) = log(6/3) = log(2)

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3 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

Explanation:-

Step 1:-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

Step 2:-

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

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$r=\sqrt{5}$

Step 3:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

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2 years ago

How to do premutations

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I would watch a bunch of videos or ask a teacher or something for help

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