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3 years ago

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I don’t understand this one

1 answer:

Masja [62]3 years ago

5 0

Answer:

see below

Step-by-step explanation:

Put -1 where x is in each expression and evaluate it.

__

You will find that the expression is zero when the numerator is zero. And you will find the numerator is zero when it has a factor that is equivalent to ...

(x +1)

Substituting x=-1 into this factor makes it be ...

(-1 +1) = 0

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Evaluating the first expression, we have ...

$\dfrac{4(x+1)}{(4x+5)}=\dfrac{4(-1+1)}{4(-1)+5}=\dfrac{4\cdot 0}{1}=0$

This first expression is one you want to "check."

You can see that the reason the expression is zero is that x+1 has a sum of zero. You can look for that same sum in the other expressions. (The tricky one is the one with the factor (x -(-1)). You know, of course, that -(-1) = +1.)

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Mary makes f dollars the first day of her new job. She makes $25 ¾ the second day of her job. In total, Mary made $60.00. Solve

Whitepunk [10]

All you have to do is: 60-25.75 = 34.25.

f = 34.25

7 0

3 years ago

Read 2 more answers

May someone please help me, please? This is about exponents by the way. About 5 questions I need help with.

Phantasy [73]

Answer:

16=16^1

1/4=(1/2)^2

25= 5^2

49=7^2

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Step-by-step explanation:

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3 years ago

#1. Graph the following equation.<br> Y=6/5x + 1

TEA [102]

Answer:

6x + 1

y= 1_ 6x

it will help you

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3 years ago

A box contains 20 miniature golf putt-putt balls; 10 are

stellarik [79]

Answer: The answer is provided below

Step-by-step explanation:

From the question, a box contains 20 miniature golf putt-putt balls out of which 10 are yellow, 6 are red, and the remaining 4 are pink.

The probability of picking a red golf ball will be:

= 6/20 × 100

= 600/20

= 30%

If the red ball is replaced, the probability of picking a yellow golf ball will be:

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5 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