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Volgvan
3 years ago
6

A wire 390 in. long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures ha

ve the same area, what are the lengths of the two pieces of wire (to the nearest tenth of an inch)?

Mathematics
2 answers:
kupik [55]3 years ago
7 0

Step-by-step explanation:

Below is an attachment containing the solution.

jeka57 [31]3 years ago
3 0

Answer:

Step-by-step explanation:

Let x is the length of the square side.

We know: the two figures have the same area

<=> x^{2} = πr^{2} (r being radius)

<=> x = \sqrt{II} r

perimeter square = 4x = 4*r*\sqrt{II}

perimeter circle = 2*r*π

<=> 390 = 4*r*\sqrt{II}  + 2*r*π

<=> r = 29.16

=> perimeter square =  4*29.16*π

=> perimeter circle = 58.32π

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A man buys 6 ½ of petrol daily. How many litres of petrol does he purchase in six days​
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6 0
2 years ago
Hey can you please help me with 8 and 10 :)
insens350 [35]
8) So remember the sum of all angles inside a triangle is equal to 180°. We first want to figure out the angle below the 110°.

A straight line is 180° so if we do 180 - 110 we get 70° therefore the angle below the 110° is 70°.

So we now know two angles in the triangle, 70° and 80°. Add these and we get 150°. Now we can determine the final angle (the one labelled (2x - 6)) which will be 30°.

So let’s make it into an equation:

2x - 6 = 30

Add 6 to both sides to isolate 2x:

2x = 36

Divide by 2 to get the value of x:

x = 18



10) Since the sum of all angles in a triangle is 180, and there are only 3 angles, we can determine A easily.

B = 10°
C = 160°

So to determine A:

180 - 160 - 10

Which gives us 10

Therefore A = 10°.


So the answers:

8) x = 18
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I hope this helps you!!
4 0
2 years ago
Help please!posted picture of question
Evgen [1.6K]
The region is in the first quadrant, and the axis are continuous lines, then x>=0 and y>=0
The region from x=0 to x=1 is below a dashed line that goes through the points:
P1=(0,2)=(x1,y1)→x1=0, y1=2
P2=(1,3)=(x2,y2)→x2=1, y2=3
We can find the equation of this line using the point-slope equation:
y-y1=m(x-x1)
m=(y2-y1)/(x2-x1)
m=(3-2)/(1-0)
m=1/1
m=1
y-2=1(x-0)
y-2=1(x)
y-2=x
y-2+2=x+2
y=x+2
The region is below this line, and the line is dashed, then the region from x=0 to x=1 is:
y<x+2 (Options A or B)

The region from x=2 to x=4 is below the line that goes through the points:
P2=(1,3)=(x2,y2)→x2=1, y2=3
P3=(4,0)=(x3,y3)→x3=4, y3=0
We can find the equation of this line using the point-slope equation:
y-y3=m(x-x3)
m=(y3-y2)/(x3-x2)
m=(0-3)/(4-1)
m=(-3)/3
m=-1
y-0=-1(x-4)
y=-x+4
The region is below this line, and the line is continuos, then the region from x=1 to x=4 is:
y<=-x+2 (Option B)

Answer: The system of inequalities would produce the region indicated on the graph is Option B

8 0
3 years ago
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