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kenny6666 [7]
3 years ago
10

Which statement is true about 7/10 and 2/6?

Mathematics
1 answer:
77julia77 [94]3 years ago
5 0

Answer: The third answer

Step-by-step explanation:

7/10 is 0.7 and 2/6 is 0.33, 2/6 would be less than 1/2 because 1/2 is 0.5, 0.7 is greater than that benchmark

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Yo! Solving linear equations with integers! These are just some practice exercises I need help with! So they should probably be
drek231 [11]

Answer:

Step-by-step explanation:

<h3>Solving linear equation with one variable:</h3>

   1) -4 + 3x = 4x - 8

        Add 4 to both sides

      -4 + 3x + 4 = 4x - 8 +  4

               3x     = 4x - 4

Subtract 4x from both sides,

        3x - 4x = -4x + 4x - 4

              -x   = -4

                \sf \boxed{\bf x = 4}

2) -5x - 8 = 2

Add 8 to both sides

 -5x - 8 + 8 = 2 + 8

            -5x  = 10

Divide both sides by (-5)

            \sf \dfrac{-5}{-5}x =\dfrac{10}{-5}

                 \sf \boxed{\bf x = -2}

3) 12r - 14 = 5(2-r)

    12r - 14 = 5*2 - 5*r

     12r - 14 = 10 - 5r

Add 14 to both sides

 12r - 14 + 14 = 10 - 5r + 14

             12r  = 24 - 5r

Add 5r to both sides

        12r + 5r = 24

               17r = 24

Divide both sides by 17

                 r = 24/17

4) 3x - 8 = -(17 + 2x)

  3x - 8 = -17 - 2x

Add 8 to both sides

3x - 8 + 8 = -17 - 2x + 8

           3x = -9 - 2x

Add 2x to both sides

      3x + 2x = -9

            5x  = -9

Divide both sides by 5

              \sf \dfrac{5}{5}x=\dfrac{-9}{5}\\\\\boxed{x = \dfrac{-9}{5}}

3 0
2 years ago
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A line drawn between the two pupils of the eyes defines which of the positioning lines?
user100 [1]

Answer:

your a very beautiful dumb.asss

Step-by-step explanation:

go give your self a bj

4 0
3 years ago
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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.
mezya [45]
1. Divide wire b in parts x and b-x. 

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= \frac{1}{2}sin60^{o}side*side=   \frac{1}{2} \frac{ \sqrt{3} }{2}  (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}
A=\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2}  )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

 2 \pi r=x

so r= \frac{x}{2 \pi }

Area of circle = \pi  r^{2}= \pi  ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2}  }* x^{2} = \frac{1}{4 \pi } x^{2}

4. Let f(x)=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)=\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b

x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

6. So one part is \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) } and the other part is b-\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

4 0
3 years ago
Read 2 more answers
If z=sprt5 -sprt2, then Izl=​
ASHA 777 [7]

Answer:

3

is the answer

because sprt - sprt = 0

then 5-2=3

3 0
3 years ago
For the following system, use the second equation to make a substitution for y in the first equation.
Anuta_ua [19.1K]

Answer:

2x+3x+4=6

Step-by-step explanation:

You'd plug in the y-equation which is y=3x+4 into the placement of the y in the first equation 2x+y=6,

Which would then result in 2x+3x+4=6, which is the answer

8 0
3 years ago
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