How do I tell whether the ordered pair is a solution of the given system ? {3x+y=4 and x-3y=-4

1 answer:

5 0

You plug the x and y values back into both equations. If both come out to be true statements ie. 4=4, 7=7 your answer is correct. If one or both of the equations turn into false statements, the ordered pair is not the solution to the system.

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The perimeter of a rectangle is 64 inches. Find the value of x if the width is 22 inches and tenth is 2x inches.

Dmitry [639]

Answer:

Step-by-step explanation:

Did you mean length instead of tenth? If so, 22*2=44, 64-44=20, 20/2=10. The length of the rectangle is 10 so to find the value of x, divide 10 by 2 so, 10/2=5, and x is equal to 5. : )

5 0

4 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.

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

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