Ask question

Ask question

All categories

3 years ago

10

What is the slope of a line that passes through (��4,��13) and (19,11)?

2 answers:

zysi [14]3 years ago

7 0

I cant see the points. Theres just a bunch of question marks.

marin [14]3 years ago

6 0

-2/15.... Change in rise over run... (change in y over change in x)

You might be interested in

If x is equal to –2, then find the value of 3x² + 5x

attashe74 [19]

Answer:

2

Step-by-step explanation:

3*(-2)to power of 2 + 5*-2

= 3*4 + 5*-2

= 12 + -10 = 2

5 0

3 years ago

Let's consider the time as a discrete variable with an increment of 1 minute. You arrive at a bus stop at 10 AM, knowing that th

Dafna11 [192]

Answer:

a) 2/3

b) 1/3

Step-by-step explanation:

Let X be the random event that measures the time you will have to wait.

Since time is uniformly distributed between 10 and 10:30 in intervals of 1 minute

P(n < X ≤ n+1) = 1/30 for every minute n=0,1,...29.

a)

P( X > 10) = 1 - P(X ≤ 10) = 1 - 10/30 = 2/3

b)

P(10 < X ≤ 20) = (20-10)/30 = 1/3

7 0

3 years ago

Can someone solve this please? (Radical Form)

Zinaida [17]

Step-by-step explanation:

16+48=68 that's it good luck

8 0

3 years ago

What is the inequality for -2<7+x

mario62 [17]

Answer:

x >-9

Step-by-step explanation:

3 0

4 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers