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JulijaS [17]
3 years ago
12

Write the series using summation notation. 9 + 29 + 129 + …

Mathematics
1 answer:
Sveta_85 [38]3 years ago
4 0

Step-by-step explanation:

We can see that this adds by 20, then 100, and so on. Furthermore, it starts at 9, so we can write this as

∑9+20n

The 9 is because we start with 9, and we have a multiplier of 20, so we put the 20 in there as so. On the bottom of the sum notation, we should have n=1 (as when we plug that in, it's our first number) and at the top you can put a variable k as we don't know the maximum. I don't think I can put numbers at the bottom and top of the sum using the tools I have, so that's why I wrote it out

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Point Q is located at (-4, 6). Point R is located at (8, 6) What is the distance from point Q to point R?
Stels [109]

Answer:

12 units

Step-by-step explanation:

Notice that as we go from Q to R, x (the horizontal distance) increases by 12 and y (the vertical distance) does not change.  Thus, the distance between the two points is merely the horizontal distance, 12 units.

8 0
3 years ago
Please help with this I am completely stuck on it
vaieri [72.5K]

Answer:

f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4

Step-by-step explanation:

Given:

The function, H(x)=\sqrt[3]{6x^{2}-4}

Solution 1:

Let f(x)=\sqrt[3]{x}

If f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}, then,

\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4

Solution 2:

Let f(x)=\sqrt[3]{x-4}. Then,

f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}

Similarly, there can be many solutions.

7 0
3 years ago
Original:19 new amount:30
Westkost [7]
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8 0
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Choose the most appropriate translation of the equation.Height(age) = 100A. The height of an oak tree (100) depends on the age o
Zigmanuir [339]
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Factor completely x3 + 8x2 − 3x − 24.
balu736 [363]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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