Answer:

Step-by-step explanation:
In this problem, one is given a circle with two secants (that is a line that intersects a circle at two points). One is given certain measurements, the problem asks one to find the unknown measurements.
The product of the lengths theorem gives a ratio between the lengths in the secants. Call the part of the secant that is inside the circle (inside), and the part of the secant between the exterior of the circle and the point of intersection of the secants (outside). The sum of (inside) and (outside) make up the entire secant, call this measurement (total). Remember, there are two secants, (
) and (
) in this situation. With these naming in mind, one can state the product of the length ratio as the following:

Alternatively, one can state it like the following ratio:

Apply this ratio to the given problem, substitute the lengths of the sides of the secants in and solve for the unknown.


Cross products, multiply the numerator and denominators of opposite sides of the fraction together,


Simplify,


Inverse operations,


Substitute this value into the equation given for the measure of (EF),

Answer:
ok. this my homework soon as i get the Q u can get it
Step-by-step explanation:
Answer:
<em>P=log(2)+log(3)+4.log(x)</em>
Step-by-step explanation:
To solve the problem, we need to recall the following properties of the logarithms:


We have the expression:

Factor 6=2*3:

Apply the property of the product:

Now apply the property of the exponent:
P=log(2)+log(3)+4.log(x)
So all you do is you divide 2 1/2 by .25 and you get 10 then you know one fourth or 25% of that package is 10 berries so the rest or 30% left over is 30 berries.
<span>(a.)
Let's say α is the angle that subtends from the top of the screen to horizontal eye-level.
Let β be the angle that subtends from the bottom of the screen to horizontal eye-level.
tanα = (22 + 10 - 4) / x = 28/x
α = arctan(28/x)
tanβ = (10 - 4) / x = 6/x
β = arctan(6/x)
Ɵ = α - β
Ɵ = arctan(28/x) - arctan(6/x)
(b.)
tanƟ = tan(α - β) = (tanα - tanβ) / (1 + tanα tanβ)
tanƟ = (28/x - 6/x) / [1 + (28/x)(6/x)]
tanƟ = (22/x) / [1 + (168/x²)]
tanƟ = 22x / (x² + 168)
Ɵ = arctan[22x / (x² + 168)]</span>