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VladimirAG [237]
3 years ago
10

Plz help me with this

Mathematics
2 answers:
madreJ [45]3 years ago
6 0

In order a0->a5: 1/3, 1, 3, 9, 27, 81

All you're doing is plugging the x-value in. So for a2, x=1, so y=3¹ which means y=3.

Andrews [41]3 years ago
4 0

Answer:   \bold{\dfrac{1}{3}}, 1, 3, 9, 27, 81

<u>Step-by-step explanation:</u>

\left\begin{array}{c|l}\underline{\quad x\quad} &\underline{3^x=y}\\-1&3^{-1}=\frac{1}{3}\\0&3^0=1\\1&3^1=3\\2&3^2=9\\3&3^3=27\\4&3^4=81\end{array}\right

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Answer:

<h2>\boxed{\bold{x \ = \ 2}}</h2>

Explanation:

3(x + 6) = 24

  • <em>Simplify both sides of the equation</em>

3(x + 6) = 24

(3)(x) + (3)(6) = 24 (Distribute)

3x + 18 = 24

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If f(x) = 2x2 - 5 and g(x) = 3x + 3, find (f - g)(x).
Vera_Pavlovna [14]
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7 0
3 years ago
How many positive four-digit integers are multiples of 5 and less than 4,000
Volgvan

Answer:

  Number of positive four-digit integers which are multiples of 5 and less than 4,000 = 600

Explanation:

 Lowest four digit positive integer = 1000

 Highest four digit positive integer less than 4000 = 3999

 We know that multiples of 5 end with 0 or 5 in their last digit.

 So, lowest four digit positive integer which is a multiple of 5 = 1000

 Highest four digit positive integer less than 4000 which is a multiple of 5  = 3995.

 So, the numbers goes like,

       1000, 1005, 1010 .....................................................3990, 3995

 These numbers are in arithmetic progression, so we have first term = 1000 and common difference = 5 and nth term(An) = 3995, we need to find n.

          An = a + (n-1)d  

           3995 = 1000 + (n-1)x 5

           (n-1) x 5 = 2995

           (n-1) = 599

            n = 600

 So, number of positive four-digit integers which are multiples of 5 and less than 4,000 = 600

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