Vertex coordinates: (h, k)
Vertex form: y = a(x-h)^2 + k
y = 2(x+3) + 2(x+4)
Use distributive property:
y = 2((x+3)+(x+4))
Simplify:
y = 2(2x+7)
y = 4x+14
This is slope - intercept form, not vertex form. Vertex form is for quadratic equations - this is a linear equation.
Answer (in slope - intercept form):
y = 4x+14
Answer:
Step-by-step explanation:
-5x + 10 > -15 is your original equation.
You need to isolate the x variable.
Start by subtracting 10.
-5x > -25
x > 5
Answer:
sec theta = (sqrt24/5) cos theta = -2/5 tan theta = (-[sqrt 21]/2) sec theta = 5/2 csc theta = (5sqrt21)/21 cot theta = (-2sqrt21)/21
Step-by-step explanation:
During the problem, secx = -5/2, we can assume that as cos = -2/5. -2 = x. 5 = r. find for Y with: x^2+y^2=r^2. After that, plug in for the variables and you get all the answers. Rationalize the square roots, don't forget.
Answer:
A, it starts at 3 since it is non proportional as well as being a postive graph.
Step-by-step explanation:
Answer:
It can be concluded that the intersection of a chord and the radius that bisects it is at right angle. The two are perpendicular.
Step-by-step explanation:
i. Construct the required circle of any radius as given in the question, then locate the chord. A chord joins two points on the circumference of a circle, but not passing through its center.
ii. Construct the radius to bisect the chord, dividing it into two equal parts.
Then it would be observed that the intersection of a chord and the radius that bisects it is at right angle. Thus, the chord and radius are are perpendicular to each other.
The construction to the question is herewith attached to this answer for more clarifications.
