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2 years ago

Giving brainliest to whoever can give me the right answers :)

Mathematics

1 answer:

vredina [299]2 years ago

8 0

Answer:

The First problem is empty

The second is (-oo,3) U [6,oo)

Step-by-step explanation:

Set C is defined for all reals less than or equal to. 3.

Set D is defined for all reals greater than or equal to 6.

The First cardinal want us to find when the two Set intersects. They never intersect so that the Set is empty.

The next cardinal want us to find when the Set contains both of their subsets.

Set C can be from any negative number up to 3 inclusive.

Set D start at 6, exclusively and continues forever to thr positve. So

the Set C And Set D will be

(-oo, 3) U [6,oo)

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the system of equatuions 2y = 14 - 2x and y= -x+7 is graphed. what is the solution to the system of equations

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Answer:

the solution is the whole line y = -x + 7

Step-by-step explanation:

both equations are identical

2y = 14 - 2x

=> y = 7 - x

which is exactly the same information as y=-x+7

so, there is no single crossing point. they are exactly covering each other. what "remains" is simply the whole line.

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3 years ago

What is (6-6)+7x7+4= ?

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<h2>Anser</h2>

Step-by-step explanation:

(6-6)+7×7+4

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4 years ago

Read 2 more answers

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)

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Answer: yes.

Step-by-step explanation:

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