Answer:
a = 28.14 (Estimated)
Step-by-step explanation:
7a + 50 = 247
=> 7a = 197
=> a = 28.1428571429 (Using Calculator)
=> a = 28.14 (Estimated)
Hoped this helped.
Answer:
You can use either of the following to find "a":
- Pythagorean theorem
- Law of Cosines
Step-by-step explanation:
It looks like you have an isosceles trapezoid with one base 12.6 ft and a height of 15 ft.
I find it reasonably convenient to find the length of x using the sine of the 70° angle:
x = (15 ft)/sin(70°)
x ≈ 15.96 ft
That is not what you asked, but this value is sufficiently different from what is marked on your diagram, that I thought it might be helpful.
__
Consider the diagram below. The relation between DE and AE can be written as ...
DE/AE = tan(70°)
AE = DE/tan(70°) = DE·tan(20°)
AE = 15·tan(20°) ≈ 5.459554
Then the length EC is ...
EC = AC - AE
EC = 6.3 - DE·tan(20°) ≈ 0.840446
Now, we can find DC using the Pythagorean theorem:
DC² = DE² + EC²
DC = √(15² +0.840446²) ≈ 15.023527
a ≈ 15.02 ft
_____
You can also make use of the Law of Cosines and the lengths x=AD and AC to find "a". (Do not round intermediate values from calculations.)
DC² = AD² + AC² - 2·AD·AC·cos(A)
a² = x² +6.3² -2·6.3x·cos(70°) ≈ 225.70635
a = √225.70635 ≈ 15.0235 . . . feet
Answer:
∠13 ≅ ∠16 - Vertical Angles Theorem
∠10 ≅ ∠14 - corresponding angles for parallel line p and q cut by the transversal s
∠5 ≅ ∠13 - corresponding angles for
parallel lines r and s cut by
the transversal q
∠1 ≅ ∠5 - corresponding angles for
parallel lines r and s cut by
the transversal q
Step-by-step explanation:
Linear Pair Theorem won't be used. When you look at the lines on the image you see that 13 and 16 are vertical from each other making there answer the vertical angles theorem. When you look at 10 and 14 you see that they lie on p and q with s going in the center of them. When you look at 5 and 13 they lie on s and r with q going down the middle of them. With 1 and 5 they also lie on p and q but r goes down the center of them instead of s.
Differentiate both sides of the equation of the circle with respect to
, treating
as a function of
:

This gives the slope of any line tangent to the circle at the point
.
Rewriting the given line in slope-intercept form tells us its slope is

In order for this line to be tangent to the circle, it must intersect the circle at the point
such that

In the equation of the circle, we have

If
, then
, so we omit this case.
If
, then
, as expected. Therefore
is a tangent line to the circle
at the point (1, -2).