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3 years ago

URGENT, PLEASE HELP ! (2/5) - 50 POINTS - ! please no wrong answers for points. ! Which scatter plot represents the data?

Mathematics

2 answers:

julsineya [31]3 years ago

7 0

Answer:

A has the points plotted correctly

Step-by-step explanation:

We need to plot the data

A has the points plotted correctly

B has the point ( 10,5) plotted on (9,5)

C is missing (-6,-5)

D is missing (-6,-5) and has (-2,1) instead of (-2,-1)

iogann1982 [59]3 years ago

7 0

Answer:

Step-by-step explanation:

It would be very helpful to write the points individually from the data. Take the x value and place it with its corresponding y value:

(1,4) ; (2,2) ; (-2,-1) ; (-2,-6) ; (5,-4) ; (-6,-5) ; (10,5)

Now find the graph that has each of these points. You can write these down and cross them out if you find them on the graph, and once you find the graph where all of these points are crossed out, that's the correct graph.

The correct graph is A.

:Done

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Step-by-step explanation:

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2 years ago

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

3 years ago

Question:

Tamiku [17]

Answer:

Step-by-step explanation:

$3x^2 + 12x + 6 = 0\\\\a=3\,,\ b=12\,,\ c=6\\\\\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x=\dfrac{-12\pm\sqrt{12^2-4\cdot3\cdot6}}{2\cdot3}=\dfrac{-12\pm\sqrt{144-72}}{2\cdot3}=\dfrac{-12\pm\sqrt{72}}{2\cdot3}=\dfrac{-12\pm6\sqrt{2}}{6}\\\\x_1=\dfrac{6(-2+\sqrt{2})}{6}=-2+\sqrt2\\\\x_2=\dfrac{6(-2-\sqrt{2})}{6}=-2-\sqrt2$

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2 years ago