We are given with the equation y"+ 9 y = t^2 * e^3 t + 6 and asked to determine the general solution to this equation. we convert the equation into r^2 + 9 = 0. The solution to this equation is r = +/- 3i. This solution converted to trigonometric function is equivalent to <span>y</span>= <span>C1 </span>cos3t + <span>C2 </span>sin<span>3t. </span>
Answer: A 2.28
Step-by-step explanation: Divide 496 by 218.
The Equation is y=+5
Why:
You want to find the equation for a line that passes through the point (-4,5) and has a slope of .
First of all, remember what the equation of a line is:
y = mx+b
Where:
m is the slope, and
b is the y-intercept
To start, you know what m is; it's just the slope, which you said was . So you can right away fill in the equation for a line somewhat to read:
y=x+b.
Now, what about b, the y-intercept?
To find b, think about what your (x,y) point means:
(-4,5). When x of the line is -4, y of the line must be 5.
Because you said the line passes through this point, right?
Now, look at our line's equation so far: . b is what we want, the 0 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (-4,5).
So, why not plug in for x the number -4 and for y the number 5? This will allow us to solve for b for the particular line that passes through the point you gave!.
(-4,5). y=mx+b or 5=0 × -4+b, or solving for b: b=5-(0)(-4). b=5.
I did this test b4, yours is answer #number 12
Convert things to their basic forms.
<span>Remember a few identities </span>
<span>sin^2 + cos^2 = 1 so </span>
<span>sin^2 = 1 - cos^2 and </span>
<span>cos^2 = 1 - sin^2 </span>
<span>I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. </span>
<span>Follow along to see how each problem was worked out. You'll catch on to the general technique. </span>
<span>====== </span>
<span>1. sec θ sin θ </span>
<span>1/cos * sin = sin/cos = tan </span>
<span>2. cos θ tan θ </span>
<span>cos * sin/cos = sin </span>
<span>3. tan^2 θ- sec^2 θ </span>
<span>sin^2 / cos^2 - 1/cos^2 </span>
<span>(sin^2 - 1)/cos^2 </span>
<span>-(1-sin^2)/cos^2 </span>
<span>-cos^2/cos^2 </span>
<span>-1 </span>
<span>4. 1- cos^2θ </span>
<span>sin^2 </span>
<span>5. (1-cosθ)(1+cosθ) </span>
<span>Remember (a+b)(a--b) = a^2 - b^2 </span>
<span>1-cos^2 = sin^2 </span>
<span>6. (secx-1) (secx+1) </span>
<span>sec^2 -1 </span>
<span>1/cos^2 - 1 </span>
<span>1/cos^2 - cos^2/cos^2 </span>
<span>(1-cos^2)/cos^2 </span>
<span>sin^2/cos62 </span>
<span>tan^2 </span>
<span>7. (1/sin^2A)-(1/tan^2A) </span>
<span>1/sin^2 - 1/(sin^2/cos^2) </span>
<span>1/sin^2 - cos^2/sin^2 </span>
<span>(1-cos^2)/sin^2 </span>
<span>sin^2/sin^2 </span>
<span>1 </span>
<span>8. 1- (sin^2θ/tan^2θ) </span>
<span>1-sin^2/(sin^2/cos^2) </span>
<span>1 - sin^2*cos^2/sin^2 </span>
<span>1-cos^2 </span>
<span>sin^2 </span>
<span>9. (1/cos^2θ)-(1/cot^2θ) </span>
<span>1/cos^2 - 1/(cos^2/sin^2) </span>
<span>1/cos^2 - sin^2/cos^2 </span>
<span>(1-sin^2)/cos^2 </span>
<span>cos^2/cos^2 </span>
<span>1 </span>
<span>10. cosθ (secθ-cosθ) </span>
<span>cos *(1/cos - cos) </span>
<span>1-cos^2 </span>
<span>sin^2 </span>
<span>11. cos^2A (sec^2A-1) </span>
<span>cos^2 * (1/cos^2 - 1) </span>
<span>1 - cos^2 </span>
<span>sin^2 </span>
<span>12. (1-cosx)(1+secx)(cosx) </span>
<span>(1-cos)(1+1/cos)cos </span>
<span>(1-cos)(cos + 1) </span>
<span>-(cos-1)(cos+1) </span>
<span>-(cos^2 - 1) </span>
<span>-(-sin^2) </span>
<span>sin^2 </span>
<span>13. (sinxcosx)/(1-cos^2x) </span>
<span>sin*cos/sin^2 </span>
<span>cos/sin </span>
<span>cot </span>
<span>14. (tan^2θ/secθ+1) +1 </span>
<span>(sin^2/cos^2)/(1/cos) + 2 </span>
<span>sin^2/cos + 2 </span>
<span>sin*tan + 2 </span>
Answer:
yes that are similar
Step-by-step explanation:
because the angles are both 50 degrees
similarity statement:
triangle DEF= triangle JEH
hope that helps bby<3
