Answer:
the picture isnt popping up
Yes the student is correct
To find the value of q, we need to find d(-8). Put another way, we need to find the value of d(x) when x = -8





So this means q = 0. Note that -0 is just 0.
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The value of r will be a similar, but now we use f(x) this time.
Plug in x = 0





Therefore, r = 2.
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For s, we plug x = 10 into f(x)





So s = 3.
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Finally, plug x = 10 into d(x) to find the value of t





A shortcut you could have taken is to note how d(x) = -f(x), so this means
d(10) = -f(10) = -9 since f(10) = 9 was found in the previous section above.
Whichever method you use, you should find that t = -3.
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<h3>In summary:</h3><h3>q = 0</h3><h3>r = 2</h3><h3>s = 3</h3><h3>t = -3</h3>
The easiest way to do this is to start with the power of 5 that is just under 219.
125 would be the closest power of 5
Now subtract 125 from 219 = 94
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The highest power of 5 closest to 94 is 25. You have to multiply 25 * 3 to get a number close enough to 94 that leaves a remainder under 25
94 - 75 = 19
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19 - 3 * 5 = 4 Same argument as above.
219_10 = 1334_5