Dimitristhym

sec(x) = -7/3 <=> \(\frac{1}{cos(x)}=\frac{-7}{3} \) <=> \(cos(x) = \frac{-3}{7}\) 

\(cos^2(x)=\frac{9}{49}\),We know \(sin^2(x) + cos^2(x) =1\)

So \(1-cos^2(x)=\frac{49-9}{49}\)<=> \(sin^2(x)=\frac{40}{49}\) => \(sin(x)=\frac{\sqrt{40}}{7}\) because is in the second quadrant its 

positive

and \(tan(x)=\frac{sin(x)}{cos(x)}=\frac{\frac{\sqrt{40}}{7}}{\frac{-3}{7}}\)<=> \(tan(x)=-\frac{\sqrt{40}}{3}\)

Hope it helps! 

1)\(\frac{ (x + 7)}{2 }=\frac{5x}{3}\) <=> \(3 (x + 7) = 2(5x)\) <=> \(3x + 21 = 10x \) <=> \(7x=21\) <=> \(x=21/7\) <=> \(x=3\)

2)\(\frac{ 9}{(- 3x)}=\frac{12}{(x + 10)}\) <=> \( (x+10)9 = 12(-3x)\)<=> \(9x + 90 = -36x\)<=>\(-45x=90 \) <=> \(x=-2\)

3) \(\frac{ (j + 4)}{j}+2 = 2 -\frac{1}{j}\),\(j≠ 0\)<=> \(\frac{(j+4)}{j} = -\frac{1}{j}\)<=> \(j+4=-1\) <=> \(j=-5\)

4) \(\frac{8}{q }+2=\frac{(q+4)}{(q-1)}\),\(q≠0,q≠ 1\)<=> \(\frac{8+2q}{q }=\frac{(q+4)}{(q-1)}\)<=>\((q-1)(8+2q)=(q+4)(q)\)<=>\(8q+2q^2-8-2q=q^2+4q\) <=> \(q^2+2q-8=0\)<=>\((q-2)(q+4)=0 \)<=> \(q=2,q=-4\)

5) Answer \(B\) Because \(x-5\) is in the denominator and NOT \(x+1\)

Hope it helps! 

Thanks!It's not the answer i was looking for but this help me! 

Oh now i see it.It's 8:57 a.m. and im awake all night I was studying.I'm sorry!

Thanks for the observation!

\(\left |x \right | + \left |y \right | = \left |x \right | +\left |y \right |\)(1)

We know:

\(\left |x \right | + \left | y \right | \)>\(\left | x+y \right |\) (2) 

We (1)+(2) 

\(\left |2x \right | + \left | 2y\right |\) >\(\left | x \right | + \left | y \right | +\left | x+y \right |\)

We prove it!

Hope it helps! 

We make concord the fractions before was 20/24 full and after is 9/24 the difference is 20/24-9/24

= 11/24 

The answer is 11/24 after - before practice!

Hope it helps! 

2/3t - 1 < t +7 so

2/3t - t < 7 + 1

-1/3t < 8

t > -3(8) 

t > -24  

and t+7 < -2t +15 

t + 2t < 15 - 7 

3t < 8

t < 8/3 

Finally tE (-24,8/3] 

He means with all numbers i understand with 5 numbers all the posible numbers we can take.I'm sorry ! :(

First of all we eill find domain of equation. x ≠ -4, x ≠  -3 because if x = -4,-3 we will have in denominator 0 

After this we start

\((x+3)(x) = -(x+4)(9) <=>\)

\(<=> x^2+3x = -9x -36 <=> \)

\(<=> x^2 + 12x + 36 = 0 <=> \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

x= -12/2 = -6 and

\(-6≠ -3,-6≠ -4 \) 

so \(x=-6\) is the answer.

Hope it helps! 

Its 5! = 1*2*3*4*5 = 120 different numbers with 5-digit numbers!