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3 years ago

Hey guys ( or gals ) I have no idea what to do please help me.

Mathematics

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

Irrational numbers cannot be written as a fraction, or the ratio of two whole numbers.

You would have to find a GCF (greatest common factor) and multiply it out of the radical.

ex.

$\sqrt{6}$ = 2 x 2 x 2

when you have to multiplier numbers then you can combine two of them to take outside the radical, meaning 2x2x2

2 $\sqrt{2\\}$

so $\sqrt{6}$ would be closest to the positive 2 on the number line and 2 notches.

$\sqrt{11}$

is an irrational number, so it might fall into 3.3 if turned into a decimal.

Step-by-step explanation:

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If a =6, what is the value of 2a(3b + 5c)?

leva [86]

Answer:

Step-by-step explanation:

Put a = 6 to the expression 2a(3b + 5c):

(2)(6)(3b + 5c) = 12(3b + 5c) <em>use the distributive property</em>

= (12)(3b) + (12)(5c) = 36b + 60c

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3 years ago

Order the integers {-10, -25, 25, 10, -50} from least to greatest.

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Answer:

-50,-25,-10,10,25

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4 years ago

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3 years ago

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If you add Natalie's age and Fred's age, the result is 39. If you add Fred's age to 4 times Natalie's age, the result is 78.

Darina [25.2K]

Step-by-step explanation:

We can write this word problem as two variables. Let us assume that:

x = Natalie's age

y = Fred's age

The first part of the word problem is that "If you add Natalie's age and Fred's age, the result is 39." Therefore:

Natalie's age + Fred's age = 39

x + y = 39

This will be our first equation. The second equation can be derived from the statement that "If you add Fred's age to 4 times Natalie's age, the result is 78." Therefore:

(4 times Natalie's age) + Fred's age = 78

4x + y = 78

We can now form a system of equations and solve for both x and y:

$\left \{ {{x + y = 39} \atop {4x + y = 78}} \right.$

The simplest way to solve would be using the Substitution method, as seen here:

x + y = 39

y = 39 - x

4x + y = 78

4x + (39 - x) = 78

3x + 39 = 78

3x = 39

x = 13

x + y = 39

13 + y = 39

y = 26

Remember that x = Natalie's age and y = Fred's age. Therefore, Natalie's age is 13 years old and Fred's age is 26 years old.

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3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

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3 years ago