Answer:
Shape: <em>Not sure what to put</em>
Domain: ( -oo, oo )
Range: ( -oo, oo )
Locater point: ( 0, 0 )
Step-by-step explanation:
Shape - You have the shape on the graph
Domain and Range - Both go on forever since there is no hole/interruption
Locater Point: I would say origin because it is the closest thing to a vertex, which is an example of a locater point.
Hopefully this helps!
Brainliest please?
It's B
Slope is negative, the other must be positive due to m1 x m2 = - 1
m1 = -2/5
The diagram shows a gradient of -2/5. So, the
reciprocal of this valve to make - 1 is 5/2 (which is the - 2/5 flipped upside down, but positive)
-2/5 x 5/2 = - 10/10 (or - 1)
Thus, the perpendicular line to this has a gradient of 5/2.
Hope this helps!
Answer:
It would take approximately 6.50 second for the cannonball to strike the ground.
Step-by-step explanation:
Consider the provided function.
We need to find the time takes for the cannonball to strike the ground.
Substitute h(t) = 0 in above function.
Multiply both sides by 10.
For a quadratic equation of the form 
the solutions are: 
Substitute a = -49, b = 305 and c=88
Ignore the negative value of t as time can't be a negative number.
Thus,
Hence, it would take approximately 6.50 second for the cannonball to strike the ground.
Answer:
angle 1
Step-by-step explanation:
Using the trigonometric mnemonic SOH CAH TOA, we know that cos or cosine is the ratio between the adjacent side and hypotenuse side.
This means that if cos A = 0.48, A is the measure of the angle which it's relative adjacent side divided by the hypotenuse of the triangle will be around 0.48.
Let's try angle 2, cos (angle 2) = adjacent / hypotenuse = 7.8 / 8.9 = 0.876404494382 ≈ 0.87 ≠ 0.48. Since the proportions are not equal, this angle cannot be the one marked as A.
Since angle 3 is a right angle, the adjacent could be either side so it cannot be correct. Thus angle 1 is correct.
