Answer:
angle DGE= 24 Degree
arc GF= 146degree
Step-by-step explanation:
angle DGE is inscribed in arc DE and arc DE is 48 degree , as we know inscribed angle is half of the arc subyending it , so half of 48 is 24 degree
arc GF subtended the angle GEF which is equal to 73 degree , and this angle is inscribed , if we say inscribed angle is half of the arc subtending it , this means the arc is twice of inscribed angle , so twice of 73 is 146 degree so
angle DGE =24 degree &
ARC GF=146 degree
Hi there!
24. 6x² + 19x + 3
1 3 multiply 6 × 3 = 18
6 1 multiply 1 × 1 = 1
18 + 1 = 19
(x + 3)(6x + 1)
I just simply found the two factors of 6 and 3 that can add up to 19.
This way is useful when the a and c don't have much factors.
27. 5y² + 13y + 6
1 2 5 × 2 = 10
5 3 1 × 3 = 3
10 + 3 = 13
(x + 2)(5x + 3)
30. 16x² + 16x + 3
For this one, 16 has a lot of factors. So you can try "Grouping" method.
You multiply 16 and 3. 16 × 3 =48
Find the factors of 48 that can add up to 16.
Hmm.. 12 and 4 would be great.
Then rewrite the equation.
16x² + 12x + 4x + 3
We would factor 16x² + 12x and 4x + 3 separately.
16x² + 12x 4x + 3 can't be factored more.
4x(4x + 3) 1(4x + 3)
(4x + 1)(4x + 3)
So your answers are....
24. (x + 3)(6x + 1)
27. (x + 2)(5x + 3)
30. (4x + 1)(4x + 3)
Hope this helped!!
Answer is C. T=13. Hope this helps.
Answer:
7
Step-by-step explanation:
-2 x 6 = -12
-6/1 ÷ -3/1 = 2
-12 - 2 = -14
-14 ÷ -2 = 7
Answer:
Step-by-step explanation:
![L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cfrac%7B2s%2B4%7D%7B%28s-3%29%5E%7B3%7D%7D%20%5D%3D)
Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3
Translation theorem:
![L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cfrac%7B2s%2B4%7D%7B%28s-3%29%5E%7B3%7D%7D%20%5D%3De%5E%7B3t%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2%28s%2B3%29%2B4%7D%7Bs%5E%7B3%7D%7D%20%5D)
Separate the fraction in a sum:
![e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])](https://tex.z-dn.net/?f=e%5E%7B3t%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2s%2B10%7D%7Bs%5E%7B3%7D%7D%20%5D%3De%5E%7B3t%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2s%7D%7Bs%5E%7B3%7D%7D%2B%5Cfrac%7B10%7D%7Bs%5E%7B3%7D%7D%20%5D%3De%5E%7B3t%7D%20%28L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%7D%5D%2B%20L%5E%7B-1%7D%5B%5Cfrac%7B10%7D%7Bs%5E%7B3%7D%7D%5D%29)
The formula for this is:
![L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cfrac%7Bn%21%7D%7Bs%5E%7Bn%2B1%7D%7D%20%5D%3Dt%5E%7Bn%7D)
Modify the expression to match the formula.
![e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])](https://tex.z-dn.net/?f=e%5E%7B3t%7D%20%282L%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%5E%7B1%2B1%7D%7D%5D%2B%20%5Cfrac%7B10%7D%7B2%7D%20L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%2B1%7D%7D%5D%29%3De%5E%7B3t%7D%20%282L%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%5E%7B1%2B1%7D%7D%5D%2B%205%20L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%2B1%7D%7D%5D%29)
Solve
![e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )](https://tex.z-dn.net/?f=e%5E%7B3t%7D%20%282L%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%5E%7B1%2B1%7D%7D%5D%2B%205%20L%5E%7B-1%7D%5B%5Cfrac%7B2%7D%7Bs%5E%7B2%2B1%7D%7D%5D%29%3De%5E%7B3t%7D%282t%2B5t%5E%7B2%7D%20%29)
