Any value that is set and cannot change is a variable. A.True B.False

1 answer:

7 0

Thank you for posting your question here at brainly. Feel free to ask more questions.

The best and most correct answer among the choices provided by the question is B.False .

Hope my answer would be a great help for you.

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Name the value of each 2in 222,222

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Ones,tenths,hundreths, thousands,tenthousands,hundredthousands

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Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks. Last

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Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks....

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3 years ago

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Find three consecutive integers whose sum is 558

SIZIF [17.4K]

Those are 185, 186 and 187

What you do is create this system of equations:

x+y+z=558 (three integers add up to 558)
x+1=y (they are consecutive, so that is translated as a number+1)
y+1=z (same logic as before)

And then you just resolve that system.

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3 years ago

The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c

Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle. We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

$\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}$

$\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01$

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3 years ago

. Given the first term "a" and the common ratio "r", generate the next three terms in the geometric sequence: a = 12.3; r = 0.5.

IRINA_888 [86]

Answer:

i wanna say 12.8, 13.3, 13.8

im not sure though, i hope this helps

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3 years ago