Answer:
12%
Step-by-step explanation:
1 - 0.88= 0.12 x 100 = 12
1:10 does not represent the students to chaperone ratio because for every 10 students, there must be one chaperone.
From the equation, it looks like we can set up a triangle like this. One side is x and the other is x + 20 because the northbound car traveled 20 miles farther. The hypotenuse is 100 because they were 100 miles apart.
Next, we can use the Pythagorean Theorem to solve for x:
a² + b² = c²
x² + (x + 20)² = 100²
This simplifies to:
x² + x² + 40x + 400 = 10000
2x² + 40x = 9600
2x² + 40x - 9600 = 0
After that, we can divided each side by 2:
x² + 20x - 4800 = 0
Next, we have to factor:
(x - 60)(x + 80) = 0
x = 60, -80
Since we know that distance can't be negative, 60 is the valid answer in this case.
x = 60
x + 20 = 80
Using this information, we know that the westbound car traveled 60 miles and the northbound car traveled 80 miles.
R = m - v + 2, where r = faces, v = vertices, and m = edges
r = 28 - 13 + 2
r = 15 + 2
r = 17, so the first answer is correct.
7. The surface area of a cone is A = pi*r*sqrt(r^2 + H^2)
A = pi*(7)(sqrt(49 + 1849)
A = pi*(7)(43.57)
A = pi*305 = 959 m^2, so the first answer is correct.
13. The volume of the slab is V = HLW
V = (5 yards)(5 yards)(1/12 yards)
V = 25/12 cubic yards
So it costs $46.00*(25/12) = $95.83 of total concrete. The third answer is correct.
21. First, find the volume of the rectangular prism. V = HLW
V = (15 cm)(5 cm)(7 cm)
V = 525 cm^3
Next, find the volume of the pyramid. V = 1/3(BH), where H is the height of the pyramid and B is the area of the base of the pyramid. Note that B = (15 cm)(5 cm) = 75 cm^2
V = (1/3)(75 cm^2)(13 cm)
V = 325 cm^3
Add the two volumes together, the total volume is 850 cm^3. The fourth answer is correct.
22. The volume of a square pyramid is V = 1/3(S^2)(H), where S is the side length and H is the height.
V = (1/3)(20^2 in^2)(21 in)
V = 2800 in^3
Now that we know the volume of this pyramid, and that both pyramids have equal volume, we plugin our V to the equation for the volume.
2800 = (1/3)(84)(S^2)
2800 = 28S^2
100 = S^2
<span>
10 in = S, so we have a side length of 10 in, and the first answer is correct. </span>
Answer:
Therefore the value of y(1)= 0.9152.
Step-by-step explanation:
According to the Euler's method
y(x+h)≈ y(x) + hy'(x) ....(1)
Given that y(0) =3 and step size (h) = 0.2.

Putting the value of y'(x) in equation (1)

Substituting x =0 and h= 0.2
![y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]](https://tex.z-dn.net/?f=y%280%2B0.2%29%5Capprox%20y%280%29%2B0.2%5B0%5Ctimes%20y%280%29-%5Cfrac12%20%28y%280%29%29%5E2%5D)
[∵ y(0) =3 ]

Substituting x =0.2 and h= 0.2
![y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]](https://tex.z-dn.net/?f=y%280.2%2B0.2%29%5Capprox%20y%280.2%29%2B0.2%5B%280.2%29%5E2%5Ctimes%20y%280.2%29-%5Cfrac12%20%28y%280.2%29%29%5E2%5D)
![\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.4%29%5Capprox%20%202.7%2B0.2%5B%280.2%29%5E2%5Ctimes%202.7-%20%5Cfrac12%282.7%29%5E2%5D)

Substituting x =0.4 and h= 0.2
![y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]](https://tex.z-dn.net/?f=y%280.4%2B0.2%29%5Capprox%20y%280.4%29%2B0.2%5B%280.4%29%5E2%5Ctimes%20y%280.4%29-%5Cfrac12%20%28y%280.4%29%29%5E2%5D)
![\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.6%29%5Capprox%20%201.9926%2B0.2%5B%280.4%29%5E2%5Ctimes%201.9926-%20%5Cfrac12%281.9926%29%5E2%5D)

Substituting x =0.6 and h= 0.2
![y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]](https://tex.z-dn.net/?f=y%280.6%2B0.2%29%5Capprox%20y%280.6%29%2B0.2%5B%280.6%29%5E2%5Ctimes%20y%280.6%29-%5Cfrac12%20%28y%280.6%29%29%5E2%5D)
![\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%280.8%29%5Capprox%20%201.6593%2B0.2%5B%280.6%29%5E2%5Ctimes%201.6593-%20%5Cfrac12%281.6593%29%5E2%5D)

Substituting x =0.8 and h= 0.2
![y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]](https://tex.z-dn.net/?f=y%280.8%2B0.2%29%5Capprox%20y%280.8%29%2B0.2%5B%280.8%29%5E2%5Ctimes%20y%280.8%29-%5Cfrac12%20%28y%280.8%29%29%5E2%5D)
![\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]](https://tex.z-dn.net/?f=%5CRightarrow%20y%281.0%29%5Capprox%20%200.8800%2B0.2%5B%280.8%29%5E2%5Ctimes%200.8800-%20%5Cfrac12%280.8800%29%5E2%5D)

Therefore the value of y(1)= 0.9152.