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irga5000 [103]
3 years ago
7

What is the solution to the equation below log5(2x-3)=2

Mathematics
1 answer:
Komok [63]3 years ago
7 0

Answer:

Step-by-step explanation:

Use the exponential form of this log equation. Rewriting this into exponential form gives us

5^2=2x-3

5-squared is 25, so

25 = 2x - 3 and

22 = 2x so

x = 11

You might be interested in
Do these side lengths make a right triangle? 5, 12, 13
lorasvet [3.4K]

Answer:

This is a right triangle

Step-by-step explanation:

We will use the Pythagorean theorem

a^2 + b^2 = c^2

If this is true it is a right triangle

5^2 + 12^2 = 13^2

25+144 = 169

169 =169

This is true so this is a right triangle

5 0
3 years ago
Please help me on this problem I don't know how to do it.
quester [9]
1: establish the angles, basically label one x (the small one) and the other y (the bigger one)

2. supplementary angles add up to 180, so therefore you have the equation x + y = 180.

3. so y (the bigger one) is 8 less than 3 times the smaller one (x) therefore you come up with the equation y = 3x - 8

4. plug in the equation of the bigger one into the other equation

x + 3x - 8 = 180
(combine like terms)
4x - 8 = 180

5. solve the equation by adding 8 to 180, then by dividing by 4.
x should equal 47

6. plug in 47 into the x + y = 180 equation, subtract 47 from 180 and you end up with 133

i hope i explained it alright
7 0
3 years ago
Hi.
NeX [460]

Answer:

(a)

1 sig: 0.005

2 sig: 0.0048

3 sig: 0.00482

(b)

1 sig: 50

2 sig: 50.

3 sig: 50.0

(c)

1 sig: 0.0010

2 sig: 0.00098

3 sig: 0.000981

Step-by-step explanation:

Significant Figures Rules:

  1. Any non-zero digit is significant.
  2. Any trailing zeros after the decimal is significant.
  3. Any zeros between 2 significant digits are significant.
  4. Zeroes before significant numbers in the decimal place are NOT significant; they are placeholders.

(a)

0.004816 - the zeros are placeholders, so they do not count as sig figs.

(b)

50.00168 - the zeros are between 2 significant figures, so they do count as sig figs.

(c)

0.0009812 - the zeros are placeholders, so they do not count as sig figs

5 0
3 years ago
Find the value of 8*10^0
WARRIOR [948]

8*10^0

<em><u>ANYTHING to the power of 0 is ALWAYS 1. </u></em>

8*1

8 is the correct answer.

7 0
3 years ago
Read 2 more answers
Hey can you please help! posted picture of question
frosja888 [35]
The solution of the system can be x - 3y = 4 only if both the equations can be simplified to x - 3y = 4.

This will mean that both the equations will result in the same line which is x - 3y = 4 and thus have infinitely many solutions.

Second equation is:

Qx - 6y = 8 
Taking 2 common we get:

(Q/2)x - 3y = 4

Comparing this equation to x- 3y = 4, we can say that 

Q/2 = 1
So,
Q = 2

Therefore, the second equation will be:

2x - 6y = 8
5 0
3 years ago
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