Answer:
Jada should have multiplied both sides of the equation by 108.
Step-by-step explanation:
The question is incomplete. Find the complete question in the comment section.
Given the equation -4/9 = x/108, in order to determine Jada's error, we need to solve in our own way as shown:
Step 1: Multiply both sides of the equation by -9/4 as shown:
-4/9 × -9/4 = x/108 × -9/4
-36/-36 = -9x/432
1 = -9x/432
1 = -x/48
Cross multiplying
48 = -x
x = -48
It can also be solved like this:
Given -4/9 = x/108
Multiply both sides by 108 to have:
-4/9 * 108 = x/108 * 108
-4/9 * 108 = 108x/108
-432/9 = x
x = -48
Jada should have simply follow the second calculation by multiplying both sides of the equation by 108 as shown.
Answer:
7 more students need to sign up.
Step-by-step explanation:
if you subtract the number of students needed by the number of students you have, you get 7 (30-23)
Since there is replacement, drawings are independent which means that the result of one of them does not affect the others.
Probability of drawing an apple is 4/10 (4 apple out of 10 fruits)
Probability of drawing an apricot is 2/10
Probability on one happening consecutively to the other (thanks to independence) is 4/10*2/10=8/100
Which can be simplified to 2/25
Answer:
(a) true
(b) true
(c) false; {y = x, t < 1; y = 2x, t ≥ 1}
(d) false; y = 200x for .005 < |x| < 1
Step-by-step explanation:
(a) "s(t) is periodic with period T" means s(t) = s(t+nT) for any integer n. Since values of n may be of the form n = 2m for any integer m, then this also means ...
s(t) = s(t +2mt) = s(t +m(2T)) . . . for any integer m
This equation matches the form of a function periodic with period 2T.
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(b) A system being linear means the output for the sum of two inputs is the sum of the outputs from the separate inputs:
s(a) +s(b) = s(a+b) . . . . definition of linear function
Then if a=b, you have
2s(a) = s(2a)
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(c) The output from a time-shifted input will only be the time-shifted output of the unshifted input if the system is time-invariant. The problem conditions here don't require that. A system can be "linear continuous time" and still be time-varying.
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(d) A restriction on an input magnitude does not mean the same restriction applies to the output magnitude. The system may have gain, for example.
The quotient is x^3 + 4x^2 -x + 1.
Solution:
By polynomial grid division, we start by the divisor 3x + 10 placed on the column headings.
3x 10
x^3 3x^4
We know that 3x^4 must be in the top left which means that the first row entry must be x^3. So the row and column multiply to 3x^4. We use this to fill in all of the first row, multiplying x^3 by the terms of the column entries.
3x 10
x^3 3x^4 10x^3
4x^2
We now got 10x^3 though we want 22x^3. The next cubic entry must then be 12x^3 so that the overall sum is 22x^3.
3x 10
x^3 3x^4 10x^3
4x^2 12x^3
Now we have 40x^2, so the next quadratic entry must be -3x^2 so that the overall sum is 37x^2.
3x 10
x^3 3x^4 10x^3
4x^2 12x^3 40x^2
-x -3x^2 -10x
This time we have -10x, so the next linear entry must be 3x so that the overall sum is 7x.
3x 10
x^3 3x^4 10x^3
4x^2 12x^3 40x^2
-x -3x^2 -10x
1 3x 10
The bottom and final term is 10, which is our desired answer. Therefore, we can now read the quotient off the first column:
3x^4+22x^3+37x^2-7x+10 / 3x + 10 = x^3 + 4x^2 -x + 1