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Black_prince [1.1K]
4 years ago
10

Calculate for log 4 4^8

Mathematics
1 answer:
jek_recluse [69]4 years ago
7 0
So log4 (4^8) is just 8
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List a value of b that will cause 4x2 + bx + 25 = 0 to have one real solution
vlada-n [284]

Answer:

20

Step-by-step explanation:

If you take (2x+5) (2x+5) = 0

then 4x^2 + 20x +25 = 0

5 0
3 years ago
What does a double line inside the walls of a shape mean?
BlackZzzverrR [31]

Answer:

Congruent

Step-by-step explanation:

I think it means they are congruent. Or if you are doing the side and angles I think it represents sides.i hope this helps ☺️

7 0
3 years ago
Read 2 more answers
Joe’s annual income has been increasing each year by the same dollar amount the first year his income was 19,500. 8 years later
Anna007 [38]

Answer:

The question is either poorly constructed or it is a trick question. We do not know "In what year was his income 30,300" because the years correlating to his income are not disclosed.  Even assuming the question was meant to be posed as "how many years later was his income 30,300" it is still flawed.

Step-by-step explanation:

To arrive at Joe's annual income increase, first subtract 19,500 from 27,900 to get 8400, which is his overall increase over the 8 years. Now divide 8400 by 8. You will get 1050.  Beginning with his income of 27,900, add 1,050 to each successive income amount and you will get the following progression:

27,900, 28,950, 30,000, 31,050, 32,100, 33,150, 34,200, 35,250, 36,300.

Therefore, two years after his income was 27,900 it would be 30,000 and 6 years after that it would be 36,300.  It would  never be 30,300.

6 0
3 years ago
What is the Tenth COMMON FACTOR of 6 and 9? (DO NOT respond with the LCM of 6 and 9)
Damm [24]
GMC Answer : 3 because of there
4 0
3 years ago
Simplify x+2/x^2+2x-3 divide by x+2/x^2-x
ira [324]

\dfrac{\frac{x+2}{x^2+2x-3}}{\frac{x+2}{x^2-x}}

If x\neq-2, then we can immediately cancel the factors of x+2:

\dfrac{\frac1{x^2+2x-3}}{\frac1{x^2-x}}=\dfrac{x^2-x}{x^2+2x-3}

Factorize the numerator and denominator:

x^2-x=x(x-1)

x^2+2x-3=(x+3)(x-1)

Next, if x\neq1, then

\dfrac{x^2-x}{x^2+2x-3}=\dfrac{x(x-1)}{(x+3)(x-1)}=\dfrac x{x+3}

8 0
3 years ago
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