Let's start with the blue states. Out of 50 states, 28 of them were blue.
Therefore, the fraction would be 28/50. Let's reduce it.
28/50=14/25 --> these are the lowest terms for this fraction
Use a calculator to find the decimal solution --> 14/25= .56
Now, let's do the red states. Out of the 50 states, 22 of them were blue.
22/50=11/25 --> lowest terms
Use a calculator --> 11/25= .43
Answer:
69
Step-by-step explanation:
it's the holy number jesus christ can confirm
Answer:
To figure out the common denominator for these fractions, I'll first need to factor that quadratic in the denominator on the right-hand side of the rational equation. This will also allow me to find the disallowed values for this equation. Factoring gives me:
x2 – 6x + 8 = (x – 4)(x – 2)
The factors of the quadratic on the right-hand side "just so happen" to be duplicates of the other denominators. This often happens in these exercises. (So often, in fact, that if you get completely different factors, you should probably go back and check your work.)
Step-by-step explanation:
Answer:
Correct integral, third graph
Step-by-step explanation:
Assuming that your answer was 'tan³(θ)/3 + C,' you have the right integral. We would have to solve for the integral using u-substitution. Let's start.
Given : ∫ tan²(θ)sec²(θ)dθ
Applying u-substitution : u = tan(θ),
=> ∫ u²du
Apply the power rule ' ∫ xᵃdx = x^(a+1)/a+1 ' : u^(2+1)/ 2+1
Substitute back u = tan(θ) : tan^2+1(θ)/2+1
Simplify : 1/3tan³(θ)
Hence the integral ' ∫ tan²(θ)sec²(θ)dθ ' = ' 1/3tan³(θ). ' Your solution was rewritten in a different format, but it was the same answer. Now let's move on to the graphing portion. The attachment represents F(θ). f(θ) is an upward facing parabola, so your graph will be the third one.
If you divide by 8, you can put the equation into intercept form. That form is ...
... x/a + y/b = 1
where <em>a</em> and <em>b</em> are the x- and y-intercepts, respectively.
Here, your equation would be
... x/(-2) + y/(-4) = 0
The graph with those intercepts is not shown with your problem statement here. See the attachment for the graph.