The easiest way to work this is to note that the a Sine wave is zero when it starts at zero, but this one starts before x=zero. Its transformation is moved left, so its phase shift is positive.
And that is amplitude is 1/2 of its minimum to maximum value (40) which equals 20. this says that is shifted upwards by 25.
So B. is the answer.
Answer:
3. 3 : 5; 3: 5
Step-by-step explanation:
First, let's find the leght of the sides of each octagon.
![A= 18in^{2}](https://tex.z-dn.net/?f=A%3D%2018in%5E%7B2%7D)
The area of an octagon is defined by
![A=2(1+\sqrt{2})l^{2}](https://tex.z-dn.net/?f=A%3D2%281%2B%5Csqrt%7B2%7D%29l%5E%7B2%7D)
Replacing the area
![18=2(1+\sqrt{2})l^{2}\\\frac{18}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{18}{3.4} } \approx 2.3 \ in](https://tex.z-dn.net/?f=18%3D2%281%2B%5Csqrt%7B2%7D%29l%5E%7B2%7D%5C%5C%5Cfrac%7B18%7D%7B2%281%2B%5Csqrt%7B2%7D%29%7D%20%3Dl%5E%7B2%7D%5C%5Cl%3D%5Csqrt%7B%5Cfrac%7B18%7D%7B3.4%7D%20%7D%20%5Capprox%202.3%20%5C%20in)
Therefore, the side of the first octagon is 1.6 inches long.
Its perimeter is: ![P=8(2.3in)=18.4in](https://tex.z-dn.net/?f=P%3D8%282.3in%29%3D18.4in)
![A=50 in^{2}](https://tex.z-dn.net/?f=A%3D50%20in%5E%7B2%7D)
![50=2(1+\sqrt{2})l^{2}\\\frac{50}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{50}{3.4} } \approx 3.8 \ in](https://tex.z-dn.net/?f=50%3D2%281%2B%5Csqrt%7B2%7D%29l%5E%7B2%7D%5C%5C%5Cfrac%7B50%7D%7B2%281%2B%5Csqrt%7B2%7D%29%7D%20%3Dl%5E%7B2%7D%5C%5Cl%3D%5Csqrt%7B%5Cfrac%7B50%7D%7B3.4%7D%20%7D%20%5Capprox%203.8%20%5C%20in)
Therefore, the side of the second octagon is 3.8 inches long.
Its perimeter is
.
Now, let's divide to find each ratio:
(the ratio between sides).
(the ratio between perimeters).
Therefore, the closest ratio is 3. 3 : 5; 3: 5
Answer: here you go
Step-by-step explanation:
A) 2x + 4y = 14
B) 4x + y = 20
If you multiply equation A by -2, then when the equations are added, the x will be eliminated.
The solutions are:
X = 4.714285714285714
Y = 1.1428571428571428
Answer:
y<-2
Step-by-step explanation:
8y-3y+6>2y+8y+16
-3y-2y>16-6
-5y>10
y<-2