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help!! Katherine O’Donnell obtain a personal loan of $3000 at 14% for 24 months. The monthly payment is $144. What is the intere
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Answer:
Step-by-step explanation:
let side of cube=x cm
volume=x³ cm³
again side=(x-3) cm
volume=(x-3)³ cm³
x³-(x-3)³=1385
(a³-b³)=(a-b)(a²+ab+b²)
(x-x+3){x²+x(x-3)+(x-3)²}=1385
3(x^2+x²-3x+x²-6x+9)=1385
3(3x²-9x+9)=1385
9x²-27x+27=1385
9x²-27x+27-1385=0
9x²-27x-1358=0
Answer: option d. x = 3π/2
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
1. Angle PAB is 90 degrees, as it is formed from the tanget to the circle at A, and the radius drawn to A.
2. AB=BC, because tangents drawn to a circle from the same point are equal.
3. PB is Common, so by the side-side-side congruence postulate, triangles ABP and CBP are congruent.
4. So measure of m(BPA)=x/2 and m(ABP)=73/2.
5.
, x= 107 degrees.
2. AB=BC, because tangents drawn to a circle from the same point are equal.
3. PB is Common, so by the side-side-side congruence postulate, triangles ABP and CBP are congruent.
4. So measure of m(BPA)=x/2 and m(ABP)=73/2.
5.
With even just two points, you can find the equation of a line in slope-intercept form.
Slope-intercept form: where
is the slope and
is the y-intercept
<u>1) Solve for the slope (</u><u>)</u>
The equation to solve for the slope is when the two points are
and
. Plug the coordinates of these points into the equation and solve for
.
Then, plug into
.
<u>1) Solve for the y-intercept (</u><u>)</u>
Then, take any of the given points and plug it into along with the slope. Isolate
to get the y-intercept. Then, plug both m and b back into
to get your final equation.
I hope this helps!