Answer:
65º
Step-by-step explanation:
- The angle of a straight line is 180º, so ∠ABD=180º and ∠ABC=(180-6x)º
- The sum of the interior angles of a triangle is 180, so (x+40)º+(3x+10)º+(180-6x)º=180
- We can solve from there, x+40+3x+10+180-6x=180
- Combine like terms, -2x+230=180
- Subtract 230, -2x=-50
- Divide by -2, x=25
- m∠CAB=(x+40)º=(25+40)º=65º
- m∠ABC=(180-6x)º=(180-150)º=30º
- m∠BCA=(3x+10)º=(75+10)º=85º
The initial value equation is ![y = \frac 12(3 - e^{-2x})](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%2012%283%20-%20e%5E%7B-2x%7D%29)
<h3>How to solve the initial value?</h3>
The equation is given as:
![\frac{dy}{dx} +2y = 3](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%2B2y%20%3D%203)
Where
Y(0) = 1
Subtract 2y from both sides in the equation
![\frac{dy}{dx} = -2y + 3](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%20%3D%20-2y%20%2B%203)
Rewrite as:
![\frac{dy}{-2y + 3} = dx](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7B-2y%20%2B%203%7D%20%20%3D%20%20dx)
Integrate both sides of the equation
![-\frac 12\ln|-2y + 3| = x + C_o](https://tex.z-dn.net/?f=-%5Cfrac%2012%5Cln%7C-2y%20%2B%203%7C%20%3D%20x%20%2B%20C_o)
Multiply through by -2
![\ln|-2y + 3| = -2x + C_1](https://tex.z-dn.net/?f=%5Cln%7C-2y%20%2B%203%7C%20%3D%20-2x%20%2B%20C_1)
Take the exponent of both sides
![-2y + 3 = Ce^{-2x](https://tex.z-dn.net/?f=-2y%20%2B%203%20%3D%20Ce%5E%7B-2x)
Next, we solve for C under the initial condition Y(0) = 1.
This gives
![-2(1) + 3 = Ce^{-2 * 0](https://tex.z-dn.net/?f=-2%281%29%20%2B%203%20%3D%20Ce%5E%7B-2%20%2A%200)
Evaluate
-2 + 3 = C
Solve for C
C = 1
Substitute C = 1 in ![-2y + 3 = Ce^{-2x](https://tex.z-dn.net/?f=-2y%20%2B%203%20%3D%20Ce%5E%7B-2x)
![-2y + 3 = e^{-2x](https://tex.z-dn.net/?f=-2y%20%2B%203%20%3D%20e%5E%7B-2x)
Next, we solve for y
![y = \frac 12(3 - e^{-2x})](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%2012%283%20-%20e%5E%7B-2x%7D%29)
Hence, the initial value equation is ![y = \frac 12(3 - e^{-2x})](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%2012%283%20-%20e%5E%7B-2x%7D%29)
Read more about initial value at:
brainly.com/question/16945606
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Answer:
14vr
Step-by-step explanation:
7vr - (-7rv)=
Subtracting a negative is like adding
7vr + (7rv)=
Changing the order of rv to vr
7vr + (7vr)=
Combing like terms
14vr
Answer:
Yes, they are equivalent.
Answer:
2.62
Step-by-step explanation:
![log_{b} \frac{b^{2}x^{\frac{5}{2} }}{\sqrt{y}}](https://tex.z-dn.net/?f=log_%7Bb%7D%20%5Cfrac%7Bb%5E%7B2%7Dx%5E%7B%5Cfrac%7B5%7D%7B2%7D%20%7D%7D%7B%5Csqrt%7By%7D%7D)
First, write the square root as exponent.
![log_{b} \frac{b^{2}x^{\frac{5}{2} }}{y^{\frac{1}{2}}}](https://tex.z-dn.net/?f=log_%7Bb%7D%20%5Cfrac%7Bb%5E%7B2%7Dx%5E%7B%5Cfrac%7B5%7D%7B2%7D%20%7D%7D%7By%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D)
Move the denominator to the numerator and negate the exponent.
![log_{b}(b^{2}x^{\frac{5}{2}}y^{-\frac{1}{2}})](https://tex.z-dn.net/?f=log_%7Bb%7D%28b%5E%7B2%7Dx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7Dy%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%29)
Use log product property.
![log_{b}(b^{2}) + log_{b}(x^{\frac{5}{2}}) + log_{b}(y^{-\frac{1}{2}})](https://tex.z-dn.net/?f=log_%7Bb%7D%28b%5E%7B2%7D%29%20%2B%20log_%7Bb%7D%28x%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%29%20%2B%20log_%7Bb%7D%28y%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%29)
Use log exponent property.
![2 log_{b}(b) + {\frac{5}{2}}log_{b}(x) - {\frac{1}{2}}log_{b}(y)](https://tex.z-dn.net/?f=2%20log_%7Bb%7D%28b%29%20%2B%20%7B%5Cfrac%7B5%7D%7B2%7D%7Dlog_%7Bb%7D%28x%29%20-%20%7B%5Cfrac%7B1%7D%7B2%7D%7Dlog_%7Bb%7D%28y%29)
Substitute values.
![2(1) + \frac{5}{2}(0.36) - \frac{1}{2}(0.56) \\2.62](https://tex.z-dn.net/?f=2%281%29%20%2B%20%5Cfrac%7B5%7D%7B2%7D%280.36%29%20-%20%5Cfrac%7B1%7D%7B2%7D%280.56%29%20%5C%5C2.62)