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777dan777 [17]
3 years ago
11

Triangle FGH with vertices F(1,8), G(5,7), and H(2,3) in the line y=x.

Mathematics
2 answers:
olganol [36]3 years ago
6 0

Answer:

F' (-1,8), G'(-5,7), H'(-2,3)

Step-by-step explanation:

sergiy2304 [10]3 years ago
4 0
Check Hathaway they will
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The volume V (in cubic feet) of a right cylinder with a height of 6 feet and radius r (in feet) is given by V=6πr2 . Solve the f
Advocard [28]

Answer:

r=8 ft

Step-by-step explanation:

V/6π=r2

r=√v/6π

r=√1206 cubic feet/6π

r=8 ft

4 0
3 years ago
Please answer me as soon as possible
adell [148]

Answer:????

Step-by-step explanation:

8 0
2 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

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.

y'(x)= x^2y(x)-\frac12y^2(x)

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

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

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]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

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]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

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]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

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]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

4 0
3 years ago
1/6×2/3<br> show me how to do it
hodyreva [135]
Multiple 6 with 3 and 1 with 2
2/18 = 19
7 0
3 years ago
Riley is playing a game that requires rolling a number cube numbered 1-6 and spinning a spinner with three consonants and one vo
Ghella [55]

Answer:

1/8 :)

Step-by-step explanation:

1/8 is correct because 12.5*8=100

5 0
3 years ago
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