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horsena [70]
3 years ago
6

Those who have taken the Algebra EOC please answer.

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
7 0

Step-by-step explanation:

To me yes.

Algebra is mostly memorization. If you know the formulas and know how to apply it, you should do good.

I say use this. Algebra topics are like building on top each other.

Heart of Algebra:

  • Review what the purpose of x in algebra.
  • Then learn things like combining like terms, and solving for x.
  • Since you know the basics of x, you can then review linear equations( imo, a big content of algebra). Stuff like slope, different linear forms, graphs of linear equations, and mostly linear equation word problems
  • Then you can review system of equations. since you know how to manipulate linear equations, etc.

Then move on to other algebra topics dealing with algebra like

  • functions, and different types of them
  • exponents, and radicals rules
  • inequalities.
  • sequences.
  • These aren't the heart of algebra but study them they are useful and it important to know them.

Since you learned different function rules, we can move on to learning exponetial functions, graphs, and word problems.

Then finally, learn most hard thing in Algebra: Quadratics.

Pratice,practice, and practice and you will pass.

Try to memorize the formulas and know when to apply it.

Good Luck

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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
Which two expressions is equivalent to 8(5+x)
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40 + 8x

Step-by-step explanation:

Distribute the 8 by multiplying by each value inside the parentheses.

Commutative property in addition and multiplication allows the terms to be in reversed order.

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x = 16/4

x = 4

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