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Salsk061 [2.6K]
3 years ago
11

What is an equation of the line that contains the point (3 -1) and is perpendicular to the line whose equation is y=-3x+2?

Mathematics
1 answer:
Svetach [21]3 years ago
3 0
The slope of a perpendicular line is the opposite reciprocal of the original slope. So the slope is 1/3.
Answer: y = x/3 - 2

Work in attached picture.

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Pls help I literally have no clue what this means
Elina [12.6K]

Answer:

I think the answer is c -2

Step-by-step explanation:

but don't quote me on that and if it's wrong i truly am sorry

8 0
2 years ago
Read 2 more answers
QUICK I NEED IN THE NEXT COUPLE MINUTE
Bad White [126]
The answer would be d
7 0
3 years ago
Read 2 more answers
Does somebody knows this one?
Katarina [22]

Answer:

ok so first we need to find the area of the circle inside which is a=Pi*raduis^2

a=pi*1(radius is half of diamater)

a=3.14

ok then the area of the sqare including the circle is 16 so

16-3.14=12.86

this is is aprox since the area of pi is infinite and i just used 3.14

Hope This Helps!!!

8 0
3 years ago
F(x) = 2x + 5
Finger [1]

9514 1404 393

Answer:

  15. f(x) +g(x) = 7x +2

  16. f(x) -g(x) = -3x +8

  17. h(x) · g(x) = 35x -21

Step-by-step explanation:

Substitute the function definition for each function expression and simplify.

__

15. f(x) + g(x) = (2x +5) +(5x -3)

  f(x) +g(x) = 7x +2

__

16. f(x) - g(x) = (2x +5) -(5x -3)

  f(x) -g(x) = -3x +8

__

17. h(x) · g(x) = (7) · (5x -3)

  h(x) · g(x) = 35x -21

4 0
3 years ago
Evaluate using <br> Definite integrals
swat32

Since [0,4]=[0,1]\cup(1,4], we can rewrite the integral as

\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt

Both integrals are quite immediate: you only need to use the power rule

\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}

to get

\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4

Now we only need to evaluate the antiderivatives:

\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15

So, the final answer is 15.

4 0
3 years ago
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