1/5x + 3 = 10
Subtract by 3 on both sides.
1/5x = 7.
Now to isolate the variable, flip the coefficient in front of the variable and multiply both sides by it.
5(1/5x) = 7(5)
x = 7(5)
x = 35
Now to check:
1/5(35) + 3 = 10
35/5 + 3 = 10
35/5 is 7.
7 + 3 = 10
10 = 10
Answer:
y=x-5
Step-by-step explanation:
y=mx+b
-5=0(3)+b
0(3)-5=b
-5=b
y=x-5
Correct me if im wrong
(tan(<em>x</em>) + cot(<em>x</em>)) / (tan(<em>x</em>) - cot(<em>x</em>)) = (tan²(<em>x</em>) + 1) / (tan²(<em>x</em>) - 1)
… = (sin²(<em>x</em>) + cos²(<em>x</em>)) / (sin²(<em>x</em>) - cos²(<em>x</em>))
… = -1/cos(2<em>x</em>)
Then as <em>x</em> approaches <em>π</em>/2, the limit is -1/cos(2•<em>π</em>/2) = -sec(<em>π</em>) = 1.
\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.
