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mestny [16]
3 years ago
12

Hi, this isn't a math question or anything. But I have a question about high schools. So I have all A's for my first marking per

iod and one C for an enirchnment. But do you think I could get into a good high school like Central? Also, I'm not taking the PSSA or big tests. Also, I'm in 7th grade if your wondering.
Mathematics
1 answer:
beks73 [17]3 years ago
6 0

Answer:

Depends on your GPA

Step-by-step explanation:

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Answer:

Do you mean how many waves of sound and light went through the galaxy

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4 years ago
What is the height of the prism if there is 30 square centimeters and the volume is 90 cubic centimeters
sammy [17]

Answer:

the height of the prism =3cm

Step-by-step explanation:

From the question :there is 30 square centimeters, we can deduce this is the area due to the unit

the volume is 90 cubic centimeters

Volume of prism = Area of base × height

We are to determine the height

90cm³ = 30cm² × height

divide both sides by 30cm²

height = 90cm³/30cm²

height = 3cm

Therefore the height of the prism is 3centimeters

8 0
3 years ago
100-99+98-97+96-95+...+8-7+6-5+4-3+2-1
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100-99+98-97+96-95+...+8-7+6-5+4-3+2-1\\\\=(100-99)+(98-97)+(96-95)+...+(4-3)+(2-1)\\\\=\underbrace{1+1+1+...+1}_{50}=\fbox{50}

8 0
3 years ago
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X+7=4 Algebraic equation
Vaselesa [24]
The answer is x=  

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Hope I helped!

5 0
3 years ago
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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
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