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monitta
2 years ago
9

Please help with either of these questions. Put the answer in the comments please

Mathematics
1 answer:
Lena [83]2 years ago
6 0

Answer:

2) distance = 4 km

Displacement = 4 km east

3) distance = 35 km

Displacement = 5 km north

Step-by-step explanation:

2. He runs for 3 km in the east direction and then stops.. He now runs another 1 km in the same direction then stops.. Thus,

Total distance covered = 3 + 1 = 4 km

Total displacement is how far he is from his starting point = 3 + 1 = 4 km east

3) Runs 20km north and then runs 15 km back from starting point.

Thus;

Total distance covered = 20 + 15 = 35 km

Total displacement is distance from starting point = 20 - 15 = 5 km north

Subtracted because an opposite direction was taken after the 20km back to her starting point.

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2 years ago
Please help I Think I'm wrong!!!!
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4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25
jarptica [38.1K]

Answer with explanation:

\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

 =e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}

Multiplying both sides by Integrating Factor  

x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

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3 years ago
The area, A, of a square whose side length is n​
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3 years ago
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