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nika2105 [10]
3 years ago
12

What is the equation of the line that passes through (4, -1) and (-2, 3)?

Mathematics
1 answer:
lesantik [10]3 years ago
7 0

Answer:

-2/3

Step-by-step explanation:

(4, -1) and (-2, 3)

-1 +4 =3

4 - 6 = -2

-4/6 divided by 2/2 equals -2/3

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A 20 ft ladder leans against a building and reaches a point 13 ft above the ground. Find, to the nearest degree, the angle that
Snezhnost [94]
Since it forms a "right triangle" we can apply trigonometry.
sin x = oppo/hyp
sin x = 13/20
x = sin^-1(13/20)
x = sin^-1(0.65)
x = 40.5
rounded to nearest degree
x = 41 degrees


Hope this helps! :)
4 0
3 years ago
4.) Which of the following correctly describes the line y = 5?
AnnZ [28]
1. It is parallel to the x-axis.
6 0
2 years ago
Read 2 more answers
If (a,1),(-2,b).(c,-3) are points on the straight line 2x + y = 3x - y +1. Find the values of a, b and c. THANK YOU FOR THR PERS
Lisa [10]

Answer:

it’s g ! i just turned it in and it’s g :)

6 0
2 years ago
Coral beef grew 19.5 mm taller. how much did it grow in meters?
Serggg [28]

1 meter = 1,000 millimeter

We can solve this by making two proportion or millimeter over meter:  \frac{millimeter}{meter}

\frac{19.5 mm}{x meter}  = \frac{1,000 mm}{1 m}

Now you must solve for the unknown meter under the first proportion:

(1,000)(x meter) = 19.5 * 1

1,000(x meter) = 19.5

x meter = 0.0195 m

19.5 mm = 0.0195 m

Hope this helped! Let me know if you have any further questions

~ Just a girl in love with Shawn Mendes

6 0
3 years ago
The rate at which rain accumulates in a bucket is modeled by the function r given by r(t)=10t−t^2, where r(t) is measured in mil
mars1129 [50]

Answer:

36 milliliters of rain.

Step-by-step explanation:

The rate at which rain accumluated in a bucket is given by the function:

r(t)=10t-t^2

Where r(t) is measured in milliliters per minute.

We want to find the total accumulation of rain from <em>t</em> = 0 to <em>t</em> = 3.

We can use the Net Change Theorem. So, we will integrate function <em>r</em> from <em>t</em> = 0 to <em>t</em> = 3:

\displaystyle \int_0^3r(t)\, dt

Substitute:

=\displaystyle \int_0^3 10t-t^2\, dt

Integrate:

\displaystyle =5t^2-\frac{1}{3}t^3\Big|_0^3

Evaluate:

\displaystyle =(5(3)^2-\frac{1}{3}(3)^3)-(5(0)^2-\frac{1}{3}(0)^3)=36\text{ milliliters}

36 milliliters of rain accumulated in the bucket from time <em>t</em> = 0 to <em>t</em> = 3.

4 0
3 years ago
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