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pshichka [43]
3 years ago
5

Write each sentence as an equation.

Mathematics
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

  1. <em> n - 2 = 52</em>
  2. <em>x - 3 = 12</em>
  3. <em>5 + x = 50 - 10</em>
  4. <em>3 + x = 60</em>
  5. <em>x + 6 = 10 - 2</em>
  6. <em>x - 9 = 12</em>
  7. <em>x + 5 = 13</em>
  8. <em>x + 10 = 20</em>
  9. <em>5x = 25</em>
  10. <em>2x = 5 + 2</em>

<em />

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have a great day!!

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Can someone please answer this question?
MrMuchimi

Answer:

y = x - 3

Step-by-step explanation:

6 0
2 years ago
Really Easy, Help pleaseee
Dima020 [189]

1 minute = 0.25km

Step-by-step explanation:

divide both numbers that were given by 8. 8÷8 is 1 and 2÷8 is 0.25.

5 0
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Read 2 more answers
Listed below is the amount of commissions earned last month for the eight members of the sales staff at Best Electronics. Calcul
Solnce55 [7]

Answer:

0.438

Step-by-step explanation:

Given the data:

980.9, 1036.5, 1099.5, 1153.9, 1409.0, 1456.4, 1718.4, 1721.2

Coefficient of skewness:

3 * (mean - median) / standard deviation

The mean of the dataset :

Σ(X) / N ; N = sample size = 8

980.9+1036.5+1099.5+1153.9+1409.0+1456.4+1718.4+1721.2

= 10575.8 / 8

= 1321.975

Median :

980.9, 1036.5, 1099.5, 1153.9, 1409.0, 1456.4, 1718.4, 1721.2

1/2(n + 1)th term

1/2(9) = 4.5th term

(1153.9+1409.0) / 2

= 1281.45

Standard deviation:

Sqrt[Σ(X - mean)²/ (N - 1)]

Using calculator :

Standard deviation estimate of population = 277.882456

Coefficient of skewness :

3(1321.975 - 1281.45) / 277.882456

121.575 / 277.882456

= 0.4375051

= 0.4375

Using software excel :

Using excel's AVERAGE, MEDIAN and STDEV.P functions, the Coefficient of skewness can be obtained using the formula :

3 * (mean - median) / standard deviation

Coefficient of skewness obtained is 0.438

8 0
2 years ago
1. Simplify:<br>4(4y-7y^{2})-9(5y+2)<br><br>2. Simplify:<br>24 – 4(5y – 6z) + 3y – 7z
Greeley [361]

Answer:

1)-

How to solve your question

Your question is

4(4−72)−9(5+2)

4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)

Simplify

1

Rearrange terms

4(4−72)−9(5+2)

4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)

4(−72+4)−9(5+2)

4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)

2

Distribute

4(−72+4)−9(5+2)

{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)

−282+16−9(5+2)

{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)

3

Distribute

−282+16−9(5+2)

-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)

−282+16−45−18

-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18

4

Combine like terms

2)

−17y+17z+24

See steps

Step by Step Solution:



STEP1:Equation at the end of step 1

((24 - 4 • (5y - 6z)) + 3y) - 7z

STEP2:

Final result :

-17y + 17z + 24

−282+16−45−18

-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18

−282−29−18

-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18

Solution

−282−29−18

4 0
2 years ago
Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7
Digiron [165]
y=x^r
\implies r(r-1)x^r+6rx^r+4x^r=0
\implies r^2+5r+4=(r+1)(r+4)=0
\implies r=-1,r=-4

so the characteristic solution is

y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}

As a guess for the particular solution, let's back up a bit. The reason the choice of y=x^r works for the characteristic solution is that, in the background, we're employing the substitution t=\ln x, so that y(x) is getting replaced with a new function z(t). Differentiating yields

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}
\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)

Now the ODE in terms of t is linear with constant coefficients, since the coefficients x^2 and x will cancel, resulting in the ODE

\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t

Of coursesin, the characteristic equation will be r^2+6r+4=0, which leads to solutions C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}, as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If z_1,z_2 are the solutions to the characteristic equation of the ODE in terms of z, then we can find another of the form z_p=u_1z_1+u_2z_2 where

u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt
u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt

where W(z_1,z_2) is the Wronskian of the two characteristic solutions. We have

u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t

u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t

\implies z_p=u_1z_1+u_2z_2
\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t

and recalling that t=\ln x\iff e^t=x, we have

\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x
4 0
2 years ago
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