Answer:
First option: cos(θ + φ) = -117/125
Step-by-step explanation:
Recall that cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)
If sin(θ) = -3/5 in Quadrant III, then cos(θ) = -4/5.
Since tan(φ) = sin(φ)/cos(φ), then sin(φ) = -7/25 and cos(φ) = 24/25 in Quadrant II.
Therefore:
cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)
cos(θ + φ) = (-4/5)(24/25) - (-3/5)(-7/25)
cos(θ + φ) = (-96/125) - (21/125)
cos(θ + φ) = -96/125 - 21/125
cos(θ + φ) = -117/125
Answer:
Yes, we can conclude that Triangle ABC is similar to triangle DEF because the measures of the 3 angles of both triangles are congruent.
Step-by-step explanation:
We have the measure of 2 angles from both triangles, and we know that triangles have 180°, so we can solve for the measure of the third angle for both triangles.
Triangle ABC:
Measure of angle A= 60°
Measure of angle C= 40°
Measure of angle B = 180°- (measure of angle A + measure of angle C) = 180° - (60° + 40°) = 80°
Triangle DEF
Measure of angle E= 80°
Measure of angle F= 40°
Measure of angle D= 180° - (measure of angle E + measure of angle F) = 180° - (80° + 40°) = 60°
The measures of the angles in Triangle ABC are: 60°, 40°, and 80°.
The measures of the angles in Triangle DEF are: 60°, 40°, and 80°.
Since the measure of 3 angles of the two triangles are the same, we know that the two triangles are similar.
Answer:
n = - 12
Step-by-step explanation:
n/9 + 2/3 = - 2/3
n/9 = - 2/3 - 2/3
n/9 = - 4/3
n = 9(- 4/3)
n = -36/3
n = - 12
Let
![y=C_1x+C_2x^3=C_1y_1+C_2y_2](https://tex.z-dn.net/?f=y%3DC_1x%2BC_2x%5E3%3DC_1y_1%2BC_2y_2)
. Then
![y_1](https://tex.z-dn.net/?f=y_1)
and
![y_2](https://tex.z-dn.net/?f=y_2)
are two fundamental, linearly independent solution that satisfy
![f(x,y_1,{y_1}',{y_1}'')=0](https://tex.z-dn.net/?f=f%28x%2Cy_1%2C%7By_1%7D%27%2C%7By_1%7D%27%27%29%3D0)
![f(x,y_2,{y_2}',{y_2}'')=0](https://tex.z-dn.net/?f=f%28x%2Cy_2%2C%7By_2%7D%27%2C%7By_2%7D%27%27%29%3D0)
Note that
![{y_1}'=1](https://tex.z-dn.net/?f=%7By_1%7D%27%3D1)
, so that
![x{y_1}'-y_1=0](https://tex.z-dn.net/?f=x%7By_1%7D%27-y_1%3D0)
. Adding
![y''](https://tex.z-dn.net/?f=y%27%27)
doesn't change this, since
![{y_1}''=0](https://tex.z-dn.net/?f=%7By_1%7D%27%27%3D0)
.
So if we suppose
![f(x,y,y',y'')=y''+xy'-y=0](https://tex.z-dn.net/?f=f%28x%2Cy%2Cy%27%2Cy%27%27%29%3Dy%27%27%2Bxy%27-y%3D0)
then substituting
![y=y_2](https://tex.z-dn.net/?f=y%3Dy_2)
would give
![6x+x(3x^2)-x^3=6x+2x^3\neq0](https://tex.z-dn.net/?f=6x%2Bx%283x%5E2%29-x%5E3%3D6x%2B2x%5E3%5Cneq0)
To make sure everything cancels out, multiply the second degree term by
![-\dfrac{x^2}3](https://tex.z-dn.net/?f=-%5Cdfrac%7Bx%5E2%7D3)
, so that
![f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y](https://tex.z-dn.net/?f=f%28x%2Cy%2Cy%27%2Cy%27%27%29%3D-%5Cdfrac%7Bx%5E2%7D3y%27%27%2Bxy%27-y)
Then if
![y=y_1+y_2](https://tex.z-dn.net/?f=y%3Dy_1%2By_2)
, we get
![-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0](https://tex.z-dn.net/?f=-%5Cdfrac%7Bx%5E2%7D3%280%2B6x%29%2Bx%281%2B3x%5E2%29-%28x%2Bx%5E3%29%3D-2x%5E3%2Bx%2B3x%5E3-x-x%5E3%3D0)
as desired. So one possible ODE would be
![-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0](https://tex.z-dn.net/?f=-%5Cdfrac%7Bx%5E2%7D3y%27%27%2Bxy%27-y%3D0%5Ciff%20x%5E2y%27%27-3xy%27%2B3y%3D0)
(See "Euler-Cauchy equation" for more info)
Answer:
Step-by-step explanation:
Givens
r = 14 miles / hour
Problem
Convert to feet per second.
Solution
Later on, you will learn how to solve these kinds of conversion problems using a method called Unit Analysis. Since you likely have not been taught this method, will just use proportions.
1 mile = 5280 feet
14 miles = x Cross multiply
x = 14 * 5280 Simplify
x = 73920 feet
1 hour = 3600 seconds
Answer
r = 73920 feet / 3600 seconds
r = 20.533 feet / secpmd