Why would I rewrite this

Mathematics

1 answer:

Ratling [72]3 years ago

7 0

You are finding 3 1/6 - 1 5/6.

To solve, you should first change it so that the numerator of the first fraction is greater than the numerator of the second. Note that both have the same denominator (6). Subtract as usual:

3 1/6 = 2 7/6

2 7/6 - 1 5/6 = 1 2/6

1 2/6 is your answer.

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O is the midpoint of segment FG. FO=3x+17, OG=7x-15. 98 points

AlladinOne [14]

O is the midpoint so both sides FO and OG need to equal each other.

3x + 17 = 7x - 15

Subtract 3x from both sides:

17 = 4x -15

Add 15 to both sides:

32 = 4x

Divide both sides by 4:

x = 32/4

X = 8

Now you have the value of X, solve for the lengths:

FO = 3x +17 = 3(8) +17 = 24 +17 = 41

OG = 7x -15 = 7(8) -15 = 56-15 = 41

Total length would be 41 + 41 = 82

5 0

3 years ago

Read 2 more answers

Pls help trig question!!

s2008m [1.1K]

$4\sin^2 x - 3 = 3 \sin x\\\\\implies 4\sin^2 x -3 \sin x -3 =0\\\\\implies 4u^2 -3u -3 =0~~~~~~~~~~~~~~;[\text{Set} ~ \sin x = u]\\\\\implies u = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4\cdot 4 \cdot (-3)}}{2(4)}~~~~~~;[\text{Apply quadratic formula}]\\\\\\\implies u = \dfrac{3\pm\sqrt{57}}8\\\\\implies \sin x = \dfrac{3\pm\sqrt{57}}8~~~~~~;[\text{Substitute back}~ u = \sin x]\\\\\text{Now,}\\\\\sin x = \dfrac{3+\sqrt{57}}8\\\\\text{No solution for}~ x\in \mathbb{R} ~\text{Since}~ -1\leq \sin x \leq 1\\$

$\sin x = \dfrac{3 - \sqrt{57}}8\\\\\implies x = n\pi + (-1)^n \sin^{-1} \left(\dfrac{3-\sqrt{57}}8\right)\\\\\\\text{ When n = 1,2 and}~ [0,2\pi]},\\\\\\x= \pi - \sin^{-1} \left(\dfrac{3-\sqrt{57}}8 \right)~ \text{and} ~x = 2\pi + \sin^{-1} \left(\dfrac{3-\sqrt{57}}8 \right)\\\\\text{In decimal,}\\\\x= 3.75~~~~ \text{and}~~~ x =5.68$