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3 years ago

Show one way to count from 82 to 512

2 answers:

6 0

Since they both end in 2 you could add 2 and then eventually get 512. Or you could do 512-82 and get 430 so then you could do 430 plus 82 is 512.

Vlad1618 [11]3 years ago

3 0

Only way is by 2 I think

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julia-pushkina [17]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Graph y=-6(3)^x please do the table and the graph 20 POINTS!

Margarita [4]

Answer:

Explanation to the answer below... (Answers in the Explanation)

Step-by-step explanation:

We Know that our rule is y = $y = -6(3)^{x}$ , which means we can use this rule to find y from x. Since we don't know what X, is, we have to assume numbers like 1, to 7. And there is no space to write on the graph, I'll put a virtual copy of a graph.

1st Point,

$y = -6(3)^{x}\\y = -6(3)^{1}\\y = -6(3)\\y = -18\\x = 1, y = -18$

2nd, point

$y = -6(3)^x\\ y = -6(3)^2\\y = -6(9)\\y = -54\\x = 2, y = -54$

3rd Point

$y = -6(3)^x\\y = -6(3)^3\\y = -6(27)\\y = -162\\x = 3, y = -162$

4th Point

$y = -6(3)^4\\y = -6(81)\\y = -486\\x = 4, y = -486\\$

5th Point

$y = -6(3)^5\\y = -6(243)\\y = -1458\\\x = 5, y = -1458$

6th Point

$y = -6(3)^x\\y = -6(3)^6\\y = -6(729)\\y = -4,374\\x = 6, y = -4,374\\$

7th Point (Last Point)

$y = -6(3)^x\\y = -6(3)^7\\y = -6(2,187)\\y. = -13,122\\x = 7, y = -13,122$

Let's now plot them on the Graph, the values are so big, the graph might look a little odd.

Always remember that we can continue the points infinitely, and there is no stop, The Graph is attached below...

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3 years ago

What is the simplest form of

ASHA 777 [7]

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step 1. the cube root is also to the 1/3 power

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step 3. a third root: 3 factors in the root comes "out" of the root as 1

step 4. 3ab*b(b)^(1/3) = 3ab^2(b)^(1/3).

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3 years ago

What is the fifth term of the sequence defined by f(n)=3(n-3)

egoroff_w [7]

$f(5)=3(5-3)=3\cdot2=6$

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3 years ago

Find Horizontal and Vertical asymptotes for y=(x^2 -9)/(4x^2 +1)

Gnom [1K]

Answer: Horizontal asymptote is $\dfrac{1}{4}$ and vertical asymptotes are $\pm \dfrac{1}{2i}$

Step-by-step explanation:

Since we have given that

$y=\dfrac{x^2-9}{4x^2+1}$

We need to find the horizontal and vertical asymptotes.

Since vertical asymptotes will occur where the denominator becomes zero.

So, here denominator is $4x^2+1$

Now,

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And the horizontal asympototes will occur when the coefficient of higher degree of numerator is divided by coefficient of higher degree of denominator.

$y=\dfrac{1}{4}$

Hence, horizontal asymptote is $\dfrac{1}{4}$ and vertical asymptotes are $\pm \dfrac{1}{2i}$

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3 years ago