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3 years ago

Can I have the answer?

Mathematics

2 answers:

dexar [7]3 years ago

7 0

Your point is (5,-3).

wlad13 [49]3 years ago

4 0

The point is correct it's (5, -3) look for the x axis first and then the number that's on the y axis

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In which quadrant is (-1/2, -1.8)?

Anna71 [15]

Answer:

quadrant 3

Step-by-step explanation:

you go to the left 1/2 and down 1.8, putting you in quadrant 3

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2 years ago

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Finds the slope of the line

Zinaida [17]

Answer:

Step-by-step explanation:

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2 years ago

40+40x0+1=<br> a.81<br> b.21<br> c.41

inn [45]

Answer:

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Step-by-step explanation:

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2 years ago

Sketch the region enclosed by y=3x and y=x^2 . Then find the area of the region.

kolezko [41]

Answer:

$A=4.5u^2$

Step-by-step explanation:

The integral of a function gives you the area under the curve, the subtraction of one of the areas from the other will give you the area in between.

The limits of integration are the points where the curves intersect each other(take the curves has a system of equation and solve for x and y):

$y=3x, y=x^2\\3x=x^2\\x^2-3x=0\\x(x-3)=0\\x_1=0\\x-3=0\\x_2=3$

$y_1=3x_1=3(0)=0\\y_2=3x_2=3(3)=9$

The integral will be the subtraction of the curve $y=3x$ and $y=x^2$ (In the graph you can see y=3x is the upper curve):

$\int\limits^3_0 {3x-x^2} \, dx =\left\frac{3x^2}{2}-\frac{x^3}{3}\right|^3_0=\frac{3(3)^2}{2}-\frac{(3)^3}{3}-\left(\frac{3(0)^2}{2}-\frac{(0)^3}{3}\right)\\\int\limits^3_0 {3x-x^2} \, dx =\frac{27}{2}-\frac{27}{3} =13.5-9=4.5u^2$

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2 years ago