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3 years ago

(8,3) and (-4,-3). Slope

Mathematics

1 answer:

Korolek [52]3 years ago

3 0

Answer:

The slope is 1/2

Step-by-step explanation:

The rate of y over the rate of x so it would be y/x. You put the equation y2-y1/x2-x1

Which would look like -3-3/-4-8= -6/-12=6/12 and then you simplify it to 1/2

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Please Help me Please please please

Vesna [10]

Answer:

The answer is,

(8+2+6) ft = 16 ft

3 0

3 years ago

I WILL MARK FIRST BRAINLIEST! I leave a thanks too! 0 = -2x^2 - 7x -1 Solve using quadratic formula PLEASEEEEEE I need the answe

slamgirl [31]

Here,

2x^2 + 7x + 1 = 0

given,

a = 2

b = 7

and c = 1

From formula,

x = (-b+-√b^2-4ac) / 2a.

We get,

x = -7/4 ± √(41)/4 Ans.

4 0

3 years ago

Bonnie is four years older than Clyde.

Doss [256]

Answer:

Bonnie's age is 18

Step-by-step explanation:

Let

x ----> Bonnie's age

y ---> Clyde's age

we know that

$x=y+4$ ----> equation A

$x-10=2(y-10)$

$x-10=2y-20$

$x=2y-20+10$

$x=2y-10$ ----> equation B

substitute equation A in equation B

$y+4=2y-10$

Solve for y

$2y-y=4+10$

$y=14$

<em>Find the value of x</em>

$x=14+4=18$

therefore

Bonnie's age is 18

Clyde's age is 14

7 0

3 years ago

In 2/3 of a hour, workers had installed 2/5 of a foor. At the rate,what fraction of the flooring is finished after 1 hour?

padilas [110]

3/5 or 60%.

2/3 hour is 40 minutes, so 1/3 is 20 minutes.

40 minutes have passed and 2/5 of the progress on the floor is complete. In 20 minutes, 1/5 of the work is complete. Add 20 minutes to 40 minutes to get 60 minutes, or an hour.
Because of this, we need to add the progress of 20 minutes and 40 minutes. 1/5+2/5=3/5.

4 0

3 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

3 years ago