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Ket [755]
3 years ago
10

Solve X/3 - 2X+1/3 = X-3/5​

Mathematics
1 answer:
iren2701 [21]3 years ago
3 0

Answer:

Answer is 3

Step-by-step explanation:

x/3-2x/1+3 =x-3/5

x/3-x/3 = x-3/5

0=x-3/5

0=x-3

3=x

sothat, x = 3 ans

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Solve the quadratic equation x2+8x−30=0 by completing the square.
diamong [38]

Answer:

               \bold{x=-4\pm\sqrt{46}}

Step-by-step explanation:

(a+b)^2=a^2+2ab+b^2\\\\\\x^2+8x-30=0\\\\\underbrace{x^2+2\cdot x\cdot 4+4^2}-4^2-30=0\\\\{}\qquad(x+4)^2\,-\,16-30=0\\\\{}\qquad\ \ (x+4)^2=46\\\\ {}\qquad\ \ x+4=\pm\sqrt{46}\\\\ {}\qquad\ \ x=-4\pm\sqrt{46}

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3 years ago
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Fiesta28 [93]

Answer:

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Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
Sabra joins a cooking club for $8.50 and pays $6.25 each month. She also joins a movie club for $12 and pays $3.75 each month. W
podryga [215]

Simplified expression for total amount spent is 10(2.05 + m)

Step-by-step explanation:

  • Step 1: Write expression for expense on joining cooking club for m months.

Expense = 8.50 + 6.25 × m

  • Step 2: Write expression for expense on joining movie club for m months.

Expense = 12 + 3.75 × m

  • Step 3: Calculate the total amount spent on both clubs

Total Amount = 8.50 + 6.25m + 12 + 3.75m = 20.5 + 10m = 10(2.05 + m)

6 0
3 years ago
Help Asap Pls Hurry I Will Give Brainliest!
dexar [7]
The answer is B
step by step explanation:
4 0
3 years ago
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The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 100
Alenkasestr [34]

Answer:

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, \mu = 310/1000 or 0.31   and  Standard deviation, \sigma = 14.63/1000 = 0.01463 .

We know that Z = \frac{X-\mu}{\sigma} ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

 P(X <=0.13) = P( \frac{X-\mu}{\sigma}  <= \frac{0.13-0.31}{0.01463} ) = P(Z <= -12.303) = P(Z > 12.303)

Since this value is not represented in the z table as the value is very high and z table is limited to x = 4.4172.

So, after seeing the table we can say that this probability is approximately less than 0.0005% .

4 0
3 years ago
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