Problem 1
Do you know what a complex number is? If you do not, you can get an answer but not one you will like much.
(3x)^2 + 27 = 0 remove the brackets. Remember to square what's inside the brackets.
9x^2 + 27 = 0 Divide both terms by 9
9x^2/9 + 27/9 = 0
x^2 + 3 = 0 Subtract 3 from both sides.
x^2 = -3 Take the square root from both sides.
x = sqrt(-3) but the square root of - 3 = 3i
x = i*sqrt(3)
Problem 2
x^ - 8x + 1 = 0
a = 1
b = - 8
c = 1
x = [- -8 +/- sqrt(b^2 - 4*a*c) ]/(2*a) Quadratic formula
x = [ 8 +/- sqrt( (-8)^2 - 4(1*1)]/2 Substitute Givens and combine
x = [ 8 +/- sqrt( 64 - 4 )] /2 Subtract 4
x = [ 8 +/- sqrt (60)]/2 Break 60 into 4 * 15
x = [ 8 +/- sqrt (4*15)]/2 Notice 4 is a perfect square. sqrt4 = 2
x = [ 8 +/- 2*sqrt(15)] / 2 Divide through by 2
x = 8/2 +/- sqrt (15)
x = 4 +/- sqrt(15) the twos were gone
Answer:
10u²
Step-by-step explanation:
The formula for area of a triangle is b*h/2. When you plug in the numbers, you get 10. Hope this helps!
Step-by-step explanation:
2x^2 + x - 28
= 2x^2 - 8x + 7x - 28
= 2x(x-4) +7(x-4)
= (2x+7)(x-4)
2x+7 = 0 or x-4 = 0
x = -7/4 or x = 4
The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
<h3>What is the derivative of the function g(x) by virtue of the Fundamental theorem of calculus as given in the task content?</h3>
g(x) = Integral; √2 ln(t) dt (with the upper and lower limits e^x and 1 respectively).
Since, it follows from the Fundamental theorem of calculus that given an integral where;
Now, g(x) = Integral f(t) dt with limits a and x, it follows that the differential of g(x);
g'(x) = f(x).
Consequently, the function g'(x) which is to be evaluated in this scenario can be determined as:
g'(x) = 
= 1
The derivative of the function g(x) as given in the task content by virtue of the Fundamental theorem of calculus is; g'(x) = √2 ln(t) dt = 1.
Read more on fundamental theorem of calculus;
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Answer:
37
Step-by-step explanation:
4 x 9 = 36
36+1 = 37
