What are the options ???
Step-by-step explanation:
Answer:
y = 1/6
Step-by-step explanation:
direct variation
y =kx
substitute in what we know to find k
1/2 = k*3
solve for k
divide by 3
1/2 /3 = k
1/6 = k
substitute k into the direct variation equation
y = 1/6 x
let x = 1
y = 1/6 *1
y = 1/6
Answer:
good news, the second one is relatively easy because it can be factored to (2x+1)(2x-3) which means that number two has solutions of -1/2 and 3/2
but for number one you have to either use the quadratic equation cause I've tried using synthetic division or just use the second equation to derive the first solutions so I tried move the graph up by two units and found that the intercepts are approximately (1+-√2)/2 or 1/2+-1/√2 for 4x^2-4x-1
Answer:
Domain: (-∞, ∞) or All Real Numbers
Range: (0, ∞)
Asymptote: y = 0
As x ⇒ -∞, f(x) ⇒ 0
As x ⇒ ∞, f(x) ⇒ ∞
Step-by-step explanation:
The domain is talking about the x values, so where is x defined on this graph? That would be from -∞ to ∞, since the graph goes infinitely in both directions.
The range is from 0 to ∞. This where all values of y are defined.
An asymptote is where the graph cannot cross a certain point/invisible line. A y = 0, this is the case because it is infinitely approaching zero, without actually crossing. At first, I thought that x = 2 would also be an asymptote, but it is not, since it is at more of an angle, and if you graphed it further, you could see that it passes through 2.
The last two questions are somewhat easy. It is basically combining the domain and range. However, I like to label the graph the picture attached to help even more.
As x ⇒ -∞, f(x) ⇒ 0
As x ⇒ ∞, f(x) ⇒ ∞
<span>The problem is to calculate the angles of the triangle. However, it is not clear which angle you have to calculate, so we are going to calculate all of them
</span>
we know that
Applying the law of cosines
c²=a²+b²-2*a*b*cos C------> cos C=[a²+b²-c²]/[2*a*b]
a=12.5
b=15
c=11
so
cos C=[a²+b²-c²]/[2*a*b]---> cos C=[12.5²+15²-11²]/[2*12.5*15]
cos C=0.694------------> C=arc cos (0.694)-----> C=46.05°-----> C=46.1°
applying the law of sines calculate angle B
15 sin B=11/sin 46.1-----> 15*sin 46.1=11*sin B----> sin B=15*sin 46.1/11
sin B=15*sin 46.1/11-----> sin B=0.9826----> B=arc sin (0.9826)
B=79.3°
calculate angle A
A+B+C=180------> A=180-B-C-----> A=180-79.3-46.1----> A=54.6°
the angles of the triangle are
A=54.6°
B=79.3°
C=46.1°
