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Goryan [66]
3 years ago
7

Any one that knows this please help thanks : )

Mathematics
2 answers:
Olin [163]3 years ago
4 0

Replace X with -2 then solve:

(-2) +3 / (-2)-1

1 / -3 which is rewritten as -1/3

Hatshy [7]3 years ago
4 0
X= -2

-2+3
--------- 1 divided by -3 = -3
-2-1

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A turtle is 20 5/6 inches below the surface of the pond. It dives to a depth of 32 1/4. What is the change of the turtles posist
BartSMP [9]

Answer:

I believe it's 11

Step-by-step explanation:

Sorry if I'm wrong

4 0
3 years ago
How would I convert -0.5x-y+5=0 into standard form with there being no coefficient for x?
juin [17]
First put the equation into standard form by isolating y:
y=-0.5x+5
To convert x to have no coefficients, the coefficient is really 1, so multiplying everything by -2 and you get
-2y=x-10
8 0
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

3 0
4 years ago
Stanley likes to go to the movie theater. Each time he goes to the movies he spends $4
Scilla [17]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2
Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

<u>Given quadratic equation</u>:

5x^2-6x-2=0

<u>Define the variables</u>:

  • a = 5
  • b = -6
  • c = -2

<u>Substitute</u> the defined variables into the quadratic formula and <u>solve for x</u>:

\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}

\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}

\implies x=\dfrac{6 \pm \sqrt{76}}{10}

\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}

\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}

\implies x=\dfrac{6 \pm2\sqrt{19}}{10}

\implies x=\dfrac{3 \pm \sqrt{19}}{5}

Therefore, the exact solutions to the given <u>quadratic equation</u> are:

x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}

Learn more about the quadratic formula here:

brainly.com/question/28105589

brainly.com/question/27953354

3 0
2 years ago
Read 2 more answers
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