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topjm [15]
3 years ago
12

PLEASE help me...! I am really stuck on this question.

Mathematics
1 answer:
Mashcka [7]3 years ago
8 0

Answer:

D

Step-by-step explanation:

the domain of 3^x  is (-∞,∞) and the range is (0,∞)

the domain of 3^x-5 is (-∞,∞) and the range is (-5,∞)

the domain doesn't change, the range (-5,∞)

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Find the value of x. 6+3x and x+14?
LenaWriter [7]
Simplifying
x + 6 = 3x + -14

Reorder the terms:
6 + x = 3x + -14

Reorder the terms:
6 + x = -14 + 3x

Solving
6 + x = -14 + 3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3x' to each side of the equation.
6 + x + -3x = -14 + 3x + -3x

Combine like terms: x + -3x = -2x
6 + -2x = -14 + 3x + -3x

Combine like terms: 3x + -3x = 0
6 + -2x = -14 + 0
6 + -2x = -14

Add '-6' to each side of the equation.
6 + -6 + -2x = -14 + -6

Combine like terms: 6 + -6 = 0
0 + -2x = -14 + -6
-2x = -14 + -6

Combine like terms: -14 + -6 = -20
-2x = -20

Divide each side by '-2'.
x = 10

Simplifying
x = 10
6 0
3 years ago
Read 2 more answers
A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of
ddd [48]

There are

\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}

ways of pairing up any 2 members from the pool of 2n contestants. Note that

(2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)

so that

\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}

4 0
2 years ago
Please help asap! Due tonight!
Contact [7]

Answer:

A, \sqrt{17}

Step-by-step explanation:

a^2 + b^2 = c^2

8^2 + b^2 = 9^2

64 + b^2 = 81

b^2 = 17

b = \sqrt{17}

4 0
3 years ago
Read 2 more answers
At which value will the graph of y=tanx have a zero? Please explain the correct answer!
Rina8888 [55]

Answer: None of the above are correct.


Step-by-step explanation:

Given function: y=\tan x

We know that the value of tan x =0 at x=0.

Also, At\ x=\frac{\pi}{3},\ \tan\frac{\pi}{3}=\sqrt{3}

At\ x=\frac{\pi}{2},\ \tan\frac{\pi}{2}=\infty

At\ x=\frac{\pi}{3},\ \tan\frac{\pi}{4}=1

Hence, the only option is correct is "None of the above are correct".

4 0
3 years ago
1. 3x + x - 2x + 8 = 3x + x
FinnZ [79.3K]

Step-by-step explanation:

1. 3x+x-2x+8=3x+x

x-2x+8=x

x+8=x+2x

8=2x

8÷2=4

x=4

2. 2(2x+4)-2x=x+18

8+4x-2x=x+18

8+2x=x+18

2x=x+10

x=10

3. 2(x+3)+3x

2(5+3)+15

16+15

31

4. 3(2x)-3x+4=2x+5

6x-3x+4=2x+5

3x+4=2x+5

3x-2x+4=5

x=1

5. 3x+2x+x=x+6

6x=x+6

5x=6

x=6/5

6. 2(x+4)+x=16

2x+8+x=16

3x+8=16

3x=8

x=8/3

Hope I calculated it right you can check it in calculator

3 0
3 years ago
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