<span>1/2x+3y/4=5 => (1/2)x + (3/4)y = 5
1/5x+3y/2=2 => (1/5)x + (3/2)y = 2
Eliminate the fractional coefficients: Mult. the first equation by 4 and mult. the second equation by 10:
2x + 15y = 20
</span><span>Multiply the 1st equation by -1:
-</span>2x - 3y = -20
2x + 15y = 20
-------------------
12y = 0, so y = 0. Then the first equation becomes 2x + 3(0) = 20, or
x=10.
Solution is (10,0).
Answer:
4z-(3z) z
Step-by-step explanation:
I hope that helps
Answer: A
Step-by-step explanation:
Letting 
,
Also, the domain of an inverse is the same as the range of the original function, so the range is 
<span>So you have composed two functions,
</span><span>h(x)=sin(x) and g(x)=arctan(x)</span>
<span>→f=h∘g</span><span>
meaning
</span><span>f(x)=h(g(x))</span>
<span>g:R→<span>[<span>−1;1</span>]</span></span>
<span>h:R→[−<span>π2</span>;<span>π2</span>]</span><span>
And since
</span><span>[−1;1]∈R→f is defined ∀x∈R</span><span>
And since arctan(x) is strictly increasing and continuous in [-1;1] ,
</span><span>h(g(]−∞;∞[))=h([−1;1])=[arctan(−1);arctan(1)]</span><span>
Meaning
</span><span>f:R→[arctan(−1);arctan(1)]=[−<span>π4</span>;<span>π4</span>]</span><span>
so there's your domain</span>
