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3 years ago

What is the slope of the line through (-9, 6) and -3, 9) and explain?

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

Answer:

m = 1 / 2

Step-by-step explanation:

The slope formula is as follows:

m = y2 - y1 / x2 - x1

Now we insert the numbers into the formula:

m = 9 - 6 / -3 - (-9)

m = 3 / 6

m = 1 / 2

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3 years ago

Solve the following equation for x. x = -2 x = 2 x = -17 x = -7 (x - 5)/2 = -6

nordsb [41]

Step-by-step explanation:

x= -2,

(x - 5) / 2= -6

( (-2) - 5) / 2= -6

(-7) / 2 = -6

-3.5 = - 6

right hand side not equal to left hand side of this equation.so,x= -2 cannot exist for this equation.

x=2,

(x - 5) / 2= -6

(2 - 5) / 2= -6

(-3) / 2= -6

-1.5 = - 6

right hand side not equal to left hand side of this equation.so,x= 2 cannot exist for this equation.

x= -17

(x - 5) / 2= -6

( ( -17) - 5) / 2= -6

(- 22) / 2= -6

-11 = -6

right hand side not equal to left hand side of this equation.so,x= -17 cannot exist for this equation.

x= -7,

(x - 5) / 2= -6

( ( -7) -5) / 2= -6

(-12) / 2= -6

-6= -6

right hand side equal to left hand side of this equation.so,x= -7 exist for this equation.

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3 years ago

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ikadub [295]

The answer should be C: 12

6 0

3 years ago

Read 2 more answers

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

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3 years ago

Read 2 more answers