Answer:
the two numbers are 16 and -25
Step-by-step explanation:
Let the two numbers be x and y
x+y=-9
x-y=-41
By substitution
y=-9-x
x-(-9-x)=-41
x+9+x=-41
2x=-41-9
2x=-50
x=-25
y=-9-(-25)
y=16
Answer: q³⁰
Explanation:
First just solve the first part using the exponent rules
p²q⁵ becomes 1/p-⁸q-²⁰ then we flip the fraction so the exponents become positive. Now we have p⁸q²⁰.
Before multiplying the other equation, we must simplify. p-⁴q⁵ becomes 1/p⁴q-⁵ and since it's the exponents being raised to a power we simply multiply the inner exponents times the outer exponent which yields 1/p⁸q-¹⁰. We must make q-¹⁰ positive so we will then bring it to the numerator of the fraction which gives us: q¹⁰/p⁸.
Multiply q¹⁰/p⁸ * p⁸q²⁰/1 = p⁸q³⁰/p⁸ divide the p exponents by each other which yields 0 since when u divide exponents you just subtract them so 8 - 8 = 0. Your answer is now q³⁰/1 or just q³⁰
\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.
Step-by-step explanation:
hope you can understand
Answer:
I think x = 1 but I’m not sure I’m sorry
Step-by-step explanation: