X^2-7x-4=0
(x-7/2)^2-(7/2)^2-4=0
(x-7/2)^2-(7)^2/(2)^2-4=0
(x-7/2)^2-49/4-4=0
(x-7/2)^2+(-49-4*4)/4=0
(x-7/2)^2+(-49-16)/4=0
(x-7/2)^2+(-65)/4=0
(x-7/2)^2-65/4=0
(x-7/2)^2-65/4+65/4=0+65/4
(x-7/2)^2=65/4
sqrt[ (x-7/2)^2 ]=+-sqrt(65/4)
x-7/2=+-sqrt(65)/sqrt(4)
x-7/2=+-sqrt(65)/2
x-7/2+7/2=+-sqrt(65)/2+7/2
x=7/2+-sqrt(65)/2
x=[7+-sqrt(65)]/2
x1=[7-sqrt(65)]/2
x2=[7+sqrt(65)]/2
Answer:
fg(-2)=f(5×-2+4)=f(-6)=8-10×-6=8+60=68
<span>(2m - n)</span>⁷ can be expanded using the binomial theorem or pascal's triangle, but writing out the work is proving to be really difficult. the application of the theorem is pretty complicated and it isn't easy to format in this textbox.
the expansion of this would be 128m⁷ - 448m⁶n + 672m⁵n² - 560m⁴n³ + 280m³n⁴ - 84m²n⁵ + 14mn⁶ - n⁷. let me know if you're genuinely lost on how to find it and i'll give a written explanation a shot, but i'd really, really encourage you to refer back to your notes if you took any in class (or the lesson, if it's an online class) and see if you can find anything which explains the binomial theorem or pascal's triangle. both of those will help you immensely.
Answer:
27 degrees and 63 degrees
Step-by-step explanation:
Let n be the number.
Angle 1: 2n-15
Angle 2: 3n
If the angles are complementary they add up to 90.
Therefore we can write an equation.
2n-15+3n=90
Combine like terms
5n-15=90
Add 15 to both sides
5n=105
n=21
Now we plug 21 in for n and solve for the angles
Angle 1 = 2(21)-15
Angle 1 = 42-15
Angle 1 = 27
Angle 2 = 3(21)
Angle 2 = 63
