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3 years ago

What is the slope of the line that passes through ( 1, -19 ) and ( -2,-7

Mathematics

1 answer:

Lapatulllka [165]3 years ago

3 0

Answer:

-4

Step-by-step explanation:

We can find the slope when given two points by using the following

m = (y2-y1)/(x2-x1)

= (-7--19)/(-2 -1)

= (-7+19)/(-2-1)

= 12/-3

=-4

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100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

dimulka [17.4K]

Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

D= (-2, 2), E = (-1, -2), F = (-4, -1)

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

Find slopes of FD and CB and compare.

m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
As we see the slopes are same

Find the slopes of FE and AB and compare.

m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
Slopes are same

Find the slopes of DE and AC and compare.

m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
Slopes are same

As per definition, midsegment is half the parallel side.

We'll show that FD = 1/2CB

FD = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2 $\sqrt{13}$
As we see FD = 1/2CB

FE = 1/2AB

FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2 $\sqrt{10}$
As we see FE = 1/2AB

DE = 1/2AC

DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2 $\sqrt{17}$
As we see DE = 1/2AC

3 0

3 years ago

Read 2 more answers

Suppose a football is kicked with an initial velocity of 82 ft/sec., at an angle of

satela [25.4K]

Answer:

The position P is:

$P = 87\^x + 75\^y$ ft Remember that the position is a vector. Observe the attached image

Step-by-step explanation:

The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity $s_0$ is:

$y(t) = y_0 + s_0t -16t ^ 2$

Where $y_0$ is the initial height = 0 for this case

We know that the initial velocity is:

82 ft/sec at an angle of 58 ° with respect to the ground.

So:

$s_0 = 82sin(58\°)$ ft/sec

$s_0 = 69.54$ ft/sec

Thus

$y(t) = 69.54t -16t ^ 2$

The height after 2 sec is:

$y(2) = 69.54 (2) -16 (2) ^ 2$

$y(2) = 75\ ft$

Then the equation that describes the horizontal position of the ball is

$X(t) = X_0 + s_0t$

Where

$X_ 0 = 0$ for this case

$s_0 = 82cos(58\°)$ ft / sec

$s_0 = 43.45$ ft/sec

$X(t) = 43.45t$

After 2 seconds the horizontal distance reached by the ball is:

$X (2) = 43.45(2)\\\\X (2) = 87\ ft$

Finally the vector position P is:

$P = 87\^x + 75\^y$ ft

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3 years ago

Please help brainliest to correct answer

jekas [21]

The first one is -4 and the second one is 6.

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3 years ago

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Can someone please help me on this question

Grace [21]

Answer: ? = 72

Step-by-step explanation:

We will set up a proportion to solve.

$\displaystyle \frac{?}{56} =\frac{45}{35}$

Now, we will cross-multiply.

? * 35 = 56 * 45

35? = 2,520

? = 72

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