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notsponge [240]
3 years ago
10

HELP IS FOR TODAY EXERCISE 2 SIMPLIFY FRACTIONS

Mathematics
1 answer:
Talja [164]3 years ago
6 0

Answer:

Do u have English version?

Step-by-step explanation:

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- x(x + 7 (x - 5)<br> Graph this
Inessa05 [86]

Answer:

=−8x2+35x

Step-by-step explanation:

multiply first

3 0
2 years ago
Solve the following differential equation: y" + y' = 8x^2
PtichkaEL [24]

Answer:

y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}

Step-by-step explanation:

Let's  find a particular solution. We need a function of the form y= ax^3+bx^2+cx+d such that

y'= 3ax^2+2bx+c and

y''= 6ax+2b

y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a =  8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is  y_p= \frac{8}{3}x^3-8x^2+16x.

Now, let's find the solution y_p of the homogeneus equation y''+y'=0 with the method of constants coefficients. Let y=e^{\lambda x}

y'=\lambda e^{\lambda x}

y''=\lambda^2e^{\lambda x}

then \lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0

e^{\lambda x}(\lambda +\lambda^2)= 0

(\lambda +\lambda^2)= 0

\lambda (1+\lambda)= 0

\lambda =0 and \lambda)= -1.

So, y_h = C_1 + C_2e^{-x} and the solution is

y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}.

5 0
3 years ago
Devon has $5,768 in his savings account, which is $2,709 more than Serena has in her savings account. How much money does Serena
Shtirlitz [24]
I got 8,477 . I hope it’s right
5 0
2 years ago
HELP PLZ! This is something I'm not very good at (that means exponents).
Marina CMI [18]

Answer:

Hey ,the last one 2*2*2*3*5*5+2*2*2*2*2=8*3*25+36=600+36=636

Hope it helps

5 0
3 years ago
3. Emma said that when you multiply three negative decimals , the product will be positive . Use these numbers to answer the que
mina [271]

Using multiplication signal rules, it is found that:

A: Emma's statement is always false.

B: The result is always negative.

C: Emma's statement is always true.

The rule used for this exercise is as follows:

  • When two numbers of different signals are multiplied, the result is negative.
  • When two numbers have the same signal, the result is positive.

Part A:

  • Three numbers are multiplied, all negative.
  • The multiplication of the first two result in a positive number.
  • Then, this positive number is multiplied by a negative number, and the result will be negative, which mean that Emma's statement is always false.

Two examples are:

-2.5(-3.8)(-2) = 9.5(-2) = -19

-2.5(-2)(-0.7) = 5(-0.7) = -3.5

Part B:

The rule is that the result is always negative.

Part C:

  • The multiplication of the first two negative numbers result in a positive number.
  • Then, this positive number is multiplied by another positive number, and the result will be positive, which mean that Emma's statement is always true.

Two examples are:

-2.5(-3.8)(2) = 9.5(2) = 19

-2.5(-2)(0.7) = 5(0.7) = 3.5

A similar problem is given at brainly.com/question/24764960

5 0
3 years ago
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