Answer:
x=-5, y=-8. (-5, -8).
Step-by-step explanation:
-x+2y=-11
5x-8y=39
---------------
5(-x+2y)=5(-11)
5x-8y=39
---------------------
-5x+10y=-55
5x-8y=39
--------------------
2y=-16
y=-16/2
y=-8
-x+2(-8)=-11
-x-16=-11
-x=-11+16
-x=5
x=-5
So, 4/3 - 2i
4/3 - 2i = 12/13 + i8/13
multiply by the conjugate:
3 + 2i/3 + 2i
= 4(3 + 2i)/(3 - 2i) (3 + 2i)
(3 - 2i) (3 + 2i) = 13
(3 - 2i) (3 + 2i)
apply complex arithmetic rule: (a + bi) (a - bi) = a^2 + b^2
a = 3, b = - 2
= 3^2 + (- 2)^2
refine: = 13
= 4(3 + 2i)/13
distribute parentheses:
a(b + c) = ab + ac
a = 4, b = 3, c = 2i
= 4(3) + 4(2i)
Simplify:
4(3) + 4(2i)
12 + 8i
4(3) + 4(2i)
Multiply the numbers: 4(3) = 12
= 12 + 2(4i)
Multiply the numbers: 4(2) = 8
= 12 + 8i
12 + 8i
= 12 + 8i/13
Group the real par, and the imaginary part of the complex numbers:
Your answer is: 12/13 + 8i/13
Hope that helps!!!
Cancel something
we cancel x's
multiply 1st equation by 5 and 2nd by 7 and add them
-35x-30y=-5
<u>35x-28y=7 +</u>
0x-58y=2
-58y=2
divide both sides by -58
y=-1/29
sub back
5x-4(-1/29)=1
5x+4/29=1
minus 4/29 from both sides
note, 1=29/29
5x=25/29
divide bot sides by 5 (or times 1/5)
x=5/29
(5/29,-1/29)
7x-99=2x+1
5x=100
x=20
mark me brainliest if the answer is correct, thank you in advance :)
Answer:
(a) f'(1)=-4
(b) y+4x-4=0
Step-by-step explanation:
<u>Tangent Line of a Function</u>
Given f(x) a real differentiable function in x=a, the slope of the tangent line of the function in x=a is given by f'(x=a). Where f' is the first derivative of f.
We are given
The derivative is
(a) The slope of the tangent line at (1,0) is
(b) The equation of the tangent line can be found with the general formula of the line:
Where m is the slope and the point (xo,yo) belongs to the line. We have m=-4, xo=1, yo=0, thus
Or, equivalently
