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ale4655 [162]
3 years ago
5

Solve the equation. x+6=x

Mathematics
2 answers:
Daniel [21]3 years ago
8 0
E. no solution. hope this helps! :)
Xelga [282]3 years ago
7 0

no solution, that means that 6=0 and that is not correct

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5.<br> If g(x) = 3x + 4, what is the value of g-1(g(-2))?
stepan [7]

Answer:

- 2

Step-by-step explanation:

The composition of inverse functions gives x

g(x) and g^{-1} (x) are inverse functions, then

g^{-1} (g(x)) = x , thus

g^{-1} (g(- 2)) = - 2

4 0
3 years ago
How to make four 2's equal 11
Ksju [112]
Answer : 2 = 4/4 + 4/4
3 0
3 years ago
Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
Coach Ellis surveyed a sample of students about their favorite sport. The circle graph below shows
CaHeK987 [17]

Answer:

C. 120 students

Step-by-step explanation:

10 + 15 + 20 +30 = 75%

100 - 75 = 25%

25% chose football which is 100 students

30% chose swimming

setup proportion:

25/100 = 30/x

25x = 100(30)

25x = 3000

x  = 120

4 0
3 years ago
How is the point (2, 9) in the scatterplot described?
AnnZ [28]
Im not sure but u can search up the name and see online i would help if i could i hope the advise i gave helps just seach up if no one can help im so sorry i wanted to help but idk this stuff yet sorry
7 0
3 years ago
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