Hello,
I am going to remember:
y'+3y=0==>y=C*e^(-3t)
y'=C'*e^(-3t)-3C*e^(-3t)
y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9
==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k
Answer:
Step-by-step explanation:
We draw the right triangle as shown in the image attached.
Δ QRS
Where R is the right angle
Given
QR = 8
QS = 17
Now, using Pythagorean Theorem, we can find RS:
Now, we know the ratio Sine as:
Where 
is the angle (here angle Q)
Opposite side is RS (which is 15)
Hypotenuse is given as QS (which is 17)
So,
This is the value of SinQ.
We don't want the angle, so we'll stop here.
See the attached picture.
<span>you are given that ABCE is an isosceles trapezoid. </span>
<span>you are given that AB is parallel to EC. </span>
<span>this means that AE is congruent to BC. </span>
<span>you are given that AE and AD are congruent. </span>
<span>triangle EAD is an isosceles triangle because AE and AD are congruent. </span>
<span>this means that angle 1 is equal to angle 3. </span>
<span>since angle 1 is equal to angle 2 and angle 3 is equal to angle 1, then angle 3 is also equal to angle 2. </span>
<span>this means that AD and BC are parellel because their corresponding angles (angles 3 and 2) are equal. </span>
<span>since AB is parallel to EC and DC is part of the same line, than AB is parallel to DC. </span>
<span>you have AB parallel to DC and AD parallel to BC. </span>
<span>if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. </span>
<span>that might be able to do it,depending on whether all these statements are acceptable without proof. </span>
<span>they are either postulates or theorems that have been previously proven. </span>
<span>if not, then you need to go a little deeper and prove some of the statements that you used.. </span>
here's my diagram.
<span>this is not a formal proof, but should give you some ideas about how to proceed. </span>
<span>you can also prove that angle 4 is equal to angle 2 because they are alternate interior angles of parallel lines. </span>
<span>you can also prove that angle 6 is equal to angle 5 because they are alternate interior angles of parallel lines. </span>
Answer:
17
Step-by-step explanation:
Answer:
![-7ab\sqrt[3]{3ab^2}](https://tex.z-dn.net/?f=-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D)
Step-by-step explanation:
Remove perfect cubes from under the radical and combine like terms.
![2ab\sqrt[3]{192ab^2}-5\sqrt[3]{81a^4b^5}=2ab\sqrt[3]{4^3\cdot 3ab^2}-5\sqrt[3]{(3ab)^3\cdot 3ab^2}\\\\=(8ab -15ab)\sqrt[3]{3ab^2}=\boxed{-7ab\sqrt[3]{3ab^2} }](https://tex.z-dn.net/?f=2ab%5Csqrt%5B3%5D%7B192ab%5E2%7D-5%5Csqrt%5B3%5D%7B81a%5E4b%5E5%7D%3D2ab%5Csqrt%5B3%5D%7B4%5E3%5Ccdot%203ab%5E2%7D-5%5Csqrt%5B3%5D%7B%283ab%29%5E3%5Ccdot%203ab%5E2%7D%5C%5C%5C%5C%3D%288ab%20-15ab%29%5Csqrt%5B3%5D%7B3ab%5E2%7D%3D%5Cboxed%7B-7ab%5Csqrt%5B3%5D%7B3ab%5E2%7D%20%7D)
