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nikitadnepr [17]
3 years ago
13

15) -1 - 2mx =6yx solve for x​

Mathematics
1 answer:
tekilochka [14]3 years ago
8 0
Answer : - 1 / 6y + 2m
The full solution is in the photo:)
( sorry if it’s wrong ;-; )

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Sophia and Olivia shared 60p so that Olivia had 10p more than Sophia. How much did Sophia have?
Taya2010 [7]

Answer:

25

Step-by-step explanation:

If Olivia had 10 more p than Sophia, then Olivia has 35, and Sophia has 25.

35 + 25 = 60!

3 0
3 years ago
8x-3(2x-4)≤ 3(x-6)
OlgaM077 [116]

The solution to the inequality expression is x ≥ 30

<h3>How to solve the inequality expression?</h3>

The inequality expression is given as:

8x - 3(2x - 4) ≤ 3(x - 6)

Open the brackets in the above inequality expression

8x - 6x + 12 ≤ 3x - 18

Collect the like terms in the above inequality expression

8x - 6x - 3x  ≤ -12 - 18

Evaluate the like terms in the above inequality expression

-x  ≤ -30

Divide both sides of the above inequality expression by -1

x ≥ 30

Hence, the solution to the inequality expression is x ≥ 30

Read more about inequality expression at

brainly.com/question/24372553


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8 0
2 years ago
Use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. (If there
True [87]

Answer:

x1 = t, x2 = -t and x3 = 0

Step-by-step explanation:

Given the system of equation

x1 + x2 + x3 = 0 .... 1

x1 + x2 + 9x3 = 0 .... 2

Subtract both equation

x3 - 9x3 = 0

-8x3 = 0

x3 = 0

Substitute x3 = 0 into equation 1

x1 + x2 + 0 = 0

x1+x2 = 0

x1 = -x2

Let t = x1

t = -x2

x2 = -t

Hence x1 = t, x2 = -t and x3 = 0

5 0
2 years ago
A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple
tester [92]

Answer:  The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation:  We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.

Therefore, the probability of event A is given by

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0
3 years ago
Select the correct answer.<br> Find the value of g(7) for the function below.<br> g(x) = zur
ale4655 [162]

Answer:

B

Step-by-step explanation:

To evaluate g(7) substitute x = 7 into g(x) , that is

g(7) = \frac{7}{8} × 7 - \frac{1}{2}

      = \frac{49}{8} - \frac{4}{8}

      = \frac{45}{8} → B

5 0
3 years ago
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