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balu736 [363]
4 years ago
5

How do I complete this formal check?

Mathematics
1 answer:
natima [27]4 years ago
8 0

Answer:

Step-by-step explanation:

The first two steps are for the purpose of eliminating fractions.  Doing so results in 4(2x - 5) = 9(x - 2), which is to be solved for x.

Perform the indicated multiplication, obtaining:

8x - 20 = 9x - 18.

Then combine like terms:  -2 = x, or

x = -2.

To complete the formal check, substitute -2 for x in the first equation.  This results in:

(1/3)(-4 - 5) = (3/4)(-2 -2), or

        -3       =        -3

... which is obviously true.  

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In a group of 50 students, 12 play basketball and volleyball
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Answer:

17 don't play any

Step-by-step explanation:

12+9+8+4=33

33-50=17

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3 years ago
Find a particular solution to y" - y + y = 2 sin(3x)
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Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
3 years ago
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3 years ago
Can anyone help me with these questions plz
KonstantinChe [14]

Answers:

  • 7:    20 miles
  • 8:    28 miles

  • 9:   1 error
  • 10:   Their speed is 50
  • 11:   365 words in 5 mins

Step-by-step explanation:

  • #7 and #8

Our equation is: f=2.25+0.20(m-1)   With f meaning "fare" (price) and m meaning "miles".

Question 7 - Juan's fare for his ride costs $6.05. We must solve for m.

Step 1. Substitute - <em>Substitute the fare for f in the equation.</em>

6.05=2.25+0.20(m-1)

Step 2. Simplify/Solve - Solve for m.

6.05=2.25+0.20(m-1)

<em>- Distribute</em>

6.05=2.25+0.2m-0.2

<em>- Subtract 0.2 from 2.25</em>

6.05=2.05+0.2m

<em>- Subtract -2.05 from 6.05</em>

4=0.2m

<em>- Divide both sides by 0.2</em>

20=m

<u>And you have your answer of 20 miles.</u>

Question 8 - Same equation, different fare.

Step 1. Substitute

7.65=2.25+0.20(m-1)

Step 2. Solve

7.65=2.25+0.20(m-1)\\7.65=2.25+0.20m-0.20\\7.65=2.05+0.20m\\5.6=0.20m\\28=m

And like so, we have m = 28

  • Questions 9 - 11

<em>The equation given is S=\frac{1}{5} (w-10e). S =  typing speed, w = words per 5 mins, and e= errors.</em>

Question 9 -

<em>We are given this information: S = 55, W  = 285. We are solving for e</em>.

Substitute -

55=\frac{1}{5} (285 -10e)

Solve -

55=\frac{1}{5} (285 -10e)\\55=\frac{1}{5} (285 -10e)\\55= 57-2e\\-2=-2e\\1=e

So they would make 1 error.

Question 10 -

<em>Information given: 300 = w, 5=e. We are solving for S</em>

Substitute -

S= \frac{1}{5} (300-10(5))

Solve -

S= \frac{1}{5} (300-10(5))\\S= \frac{1}{5} (300-50)\\S= \frac{1}{5} (300-50)\\S= \frac{1}{5} (250)\\S= 50

Their speed is 50.

Question 11 -

<em>Information given: S = 65, e = 4. We are solving for w.</em>

Substitute -

65=\frac{1}{5} (w-10(4))

Solve -

65=\frac{1}{5} (w-10(4))\\65=\frac{1}{5} (w-40)\\65=\frac{1}{5}w-8\\73=\frac{1}{5}w\\365=w

So w = 365.

  • Hoped this helped!~

7 0
4 years ago
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