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3 years ago

What is the slope of the line through (-9,6) (-6,-9)

Mathematics

1 answer:

mina [271]3 years ago

8 0

Answer:

The slope of the line through (−9,6) and (−6,−9) is −5

Step-by-step explanation:

Hope it helps

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Ben os travelling home to brimingham on the m1 and m42. both motorways have road works. He has a 50-50 chance of getting delay o

Contact [7]

The answer is 1/8 or 12.5%

8 0

3 years ago

Give the unit rate for the better deal. $156 4 g or $280 7 g

Alexus [3.1K]

Answer:

$156 for 4 g

I did not rally understand how you worded your question so sorry if I got it wrong! <3

Step-by-step explanation:

$156 for 4 g =

$156 / 4 = $39 dollars for 1 g

$280 for 7 g =

$280/7 = $40 dollars for 1 g

8 0

3 years ago

Read 2 more answers

1/2+1/4+1/8 what is this

koban [17]

Answer:

7/8

Step-by-step explanation:

1/2+1/4=3/4

3/4 + 1/8=7/8

8 0

3 years ago

Read 2 more answers

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

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

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

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

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

5 0

3 years ago

Find h(x) +g(x). h(x)=x^2+6x and g(x)=-2x-3

AURORKA [14]

Answer:

$h(x) + g(x) = x^{2} + 4x - 3$

Step-by-step explanation:

h(x) + g(x) = $(x^{2} + 6x) + (-2x - 3)$

$x^{2} + 4x - 3$

3 0

3 years ago