Answer:
and ? what is the question
Step-by-step explanation:
x = -8
Step-by-step explanation:
Step 1: Subtract 1 from both sides

Divide both sides of the equation by -10

Answer:
Between 1000 and 5000 snowboards will make the function AP(x) >0.
Step-by-step explanation:
Since x can only take possitive values, we have that AP(x) = P(x)/x > 0 if and only if P(x) > 0.
In order to find when P(x) > 0, we find the values from where it is 0 and then we use the Bolzano Theorem.
P(x) = R(x) - C(x) = -x²+10x - (4x+5) = -x²+6x - 5. the roots of P can be found using the quadratic formula:

Therefore, P(1) = P(5) = 0. Lets find intermediate values to apply Bolzano Theorem:
- P(0) = -5 < 0 ( P is negative in (-∞ , 1) )
- P(2) = -4+6*2-5 = 3 > 0 (P is positive in (1,5) )
- P(6) = -36+36-5 = -5 < 0 (P is negative in (5, +∞) )
The production levels that make AP(x) >0 are between 1000 and 5000 snowboards (because we take x by thousands)
To find the 20th term in this sequence, we can simply keep on adding the common difference all the way until we get up to the 20th term.
The common difference is the number that we are adding or subtracting to reach the next term in the sequence.
Notice that the difference between 15 and 12 is 3.
In other words, 12 + 3 = 15.
That 3 that we are adding is our common difference.
So we know that our first term is 12.
Now we can continue the sequence.
12 ⇒ <em>1st term</em>
15 ⇒ <em>2nd term</em>
18 ⇒ <em>3rd term</em>
21 ⇒ <em>4th term</em>
24 ⇒ <em>5th term</em>
27 ⇒ <em>6th term</em>
30 ⇒ <em>7th term</em>
33 ⇒ <em>8th term</em>
36 ⇒ <em>9th term</em>
39 ⇒ <em>10th term</em>
42 ⇒ <em>11th term</em>
45 ⇒ <em>12th term</em>
48 ⇒ <em>13th term</em>
51 ⇒ <em>14th term</em>
54 ⇒ <em>15th term</em>
57 ⇒ <em>16th term</em>
60 ⇒ <em>17th term</em>
63 ⇒ <em>18th term</em>
66 ⇒ <em>19th term</em>
<u>69 ⇒ </u><u><em>20th term</em></u>
<u><em></em></u>
This means that the 20th term of this arithemtic sequence is 69.
Answer:
210
Step-by-step explanation:
Here comes the problem from Combination.
We are being asked to find the number of ways out in which 3 students may sit on 7 seats in a row. Please see that in this case the even can not be repeated.
Let us start with the student one. For him all the 7 seats are available to sit. Hence number of ways for him to sit = 7
Let us see the student second. For him there are only 6 seats available to sit as one seat has already been occupied. Hence number of ways for him to sit = 6
Let us see the student third. For him there are only 5 seats available to sit as two seat has already been occupied. Hence number of ways for him to sit = 5
Hence the total number of ways for three students to be seated will be
7 x 6 x 5
=210