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3 years ago

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