$\frac{1}{2} x - \frac{3}{4} = \frac{3}{8} - \frac{5}{8} \\ \frac{1}{2} x = \frac{3}{8} - \frac{5}{8} + \frac{3}{4} \\ \frac{1}{2} x = \frac{3}{8} - \frac{5}{8} + \frac{6}{8} \\ \frac{1}{2}x = \frac{4}{8} \\ \frac{1}{2} x = \frac{1}{2} \\ x = 1$

5 0

2 years ago

PLEASE HELP I NEED TO PASS THIS EXAM! Given that triangle IHJ is equiangular, find x and y.

jok3333 [9.3K]

Use the interior angles theorem for the triangles I hope my answer helped if you have any questions don't be afraid to ask (o^▽^o) <3

8 0

2 years ago

Read 2 more answers

Find the derivative of the function. y = sin−1(6x + 1)

liraira [26]

If you don't know the derivative of the inverse of sine, you can use implicit differentiation. Apply sine to both sides:

$y=\sin^{-1}(6x+1)\implies\sin y=6x+1$

(true for y between -π/2 and π/2)

Now take the derivative of both sides and solve for it:

$\cos y\dfrac{\mathrm dy}{\mathrm dx}=6$

$\dfrac{\mathrm dy}{\mathrm dx}=6\sec y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\cos\left(\sin^{-1}(6x+1)\right)}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\sqrt{1-(6x+1)^2}}$

7 0

2 years ago