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Mrac [35]
3 years ago
9

8x-6= 2/3(6x+15)please solve for X​

Mathematics
2 answers:
77julia77 [94]3 years ago
6 0

Answer:

x=4

Step-by-step explanation:

zavuch27 [327]3 years ago
3 0

Answer:

x=4

Step-by-step explanation:

8x-6=2/3(6x+15)

8x-6= 2(6x+15)

3

8x-6=2(6x+15)

3

8x-6= 12x+30

3

8x-6=12x+30

3

8x-6+6=12x30+6

3

8x=12x+30+6

3

8x=12x+30+6

3

3.8x=3(12x+30+6)

3

24x=12x+48

24×-12x=12x48-12x

12x=48

12x=28

12x= 48

12. 12

x=4 ik its long

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PLEASE HELP ME BE CORRECT PLEASE <br><br> TELL ME WHERE to PUT EACH POINT
loris [4]

Answer:

Point A:

(3, -5)

Point B:

(6, -2)

Point C:

(5, -7)

Step-by-step explanation:

Background:

Moving to the right means adding to the x.

Moving to the left means subtracting from the x.

Moving up means adding to the y.

Moving down means subtracting from the y.

So take each point and add 3 to the x, and subtract 4 from they y.

Point B:

(3, 2) → (6, -2)

Point A:

(0, -1) → (3, -5)

Point C:

(2, -3) → (5, -7)

3 0
2 years ago
Pls HELP I BEG YOUUUU
Stells [14]

Answer:

x = 60

Step-by-step explanation:

(x+22) + (x+6) + (x-28) = 180

x + 22 + x + 6 + x - 28 = 180

x + x + x + 22 + 6 - 28 = 180

3x = 180

x = 180/3

x = 60

6 0
2 years ago
Roger has a 0.250 batting average. If he went up to bat 240 times,how many times did he fail to get any hits?( A 0.250 batting a
gulaghasi [49]

Answer:

lol i need help with the same question

Step-by-step explanation:

7 0
3 years ago
Find a square root that lies between 17 and 18.
olga nikolaevna [1]
Easy
if x is the number
sqrt x is between 17 and 18

17<√x<18
squaer the whole thing
289<x<324
x could be 300
it could be sqrt of 300
√300=10√3

one is 10√3
5 0
3 years ago
Find the range of p in equation p(x+1) (x-3)=x-4p-2 has no real roots. ​
Tanya [424]

Answer:

\displaystyle p > \frac{1}{4}.

Step-by-step explanation:

Expand the left-hand side of this equation:

p\, (x + 1)\, (x - 3) = p\, \left(x^2 - 2\, x - 3\right) = p\, x^2 - 2\, p\, x - 3\, p.

Collect the terms, so that this quadratic equation is in the form a\, x^2 + b\, x+ c = 0:

p\, x^2 - 2\, p\, x - 3\, p = x - 4\, p - 2.

p\, x^2 - (2\, p + 1)\, x + (p + 2) = 0.

In this equation:

  • a = p.
  • b = -(2\, p + 1).
  • c = p + 2.

Calculate the quadratic discriminant of this quadratic equation:

\begin{aligned}b^2 - 4\, a\, c &= (-(2\, p + 1))^2 - 4\, p\, (p + 2) \\ &= 4\, p^2 + 4\, p + 1 - 4\, p^2- 8\, p = -4\, p + 1\end{aligned}.

A quadratic equation has no real root if its quadratic discriminant is less then zero. As a result, this quadratic equation will have no real root when -4\, p + 1 < 0. Solve for the range of p:

\displaystyle p > \frac{1}{4}.

4 0
3 years ago
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