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2 years ago

8

“simplify the expression by combining like terms”

1 answer:

larisa86 [58]2 years ago

7 0

If I’m not wrong, it’d be 7x-3

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What does 1/4 subtract 1/11 equal

Anastasy [175]

Answer:

7/44

Step-by-step explanation:

you can't simply subtract 1/11 from 1/4 because the denominators are not the same . Meaning you have to convert the denominators into a similar number. Transformers in even number and a consecutive number while 11 is an odd number and a prime number they don't really agree on anything 11 can only be divided by itself and 1 wall for can be divided by a multitude of things. Because of them not exactly agreeing on any specific category , you have to multiply them by each other . So your new fractions should look like 11 / 44 and 4 / 44 . from there you can easily subtract 4 from 11 and get 7 / 44 now normally you can reduce these types of fractions but because seven can only be divided by itself and 44 is not a factor of 7 you cannot reduce this fraction .

4 0

3 years ago

Which of the following comparisons is true?

nlexa [21]

D) 436.683 < 436.684

5 0

3 years ago

((4)/(3v))-((1)/(v))=((v+2)/(2v^2))

Lerok [7]

Step-by-step explanation:

sper că te-am ajutat

5 0

2 years ago

Rewrite the expression with the given base. 25^3x with a base of 5.

svp [43]

$Use\ (a^n)^m=a^{nm}\\\\25^{3x}=(5^2)^{3x}=5^{2\cdot3x}=5^{6x}$

6 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago