George C.
Jul 24, 2018
(
x
+
2
)
(
x
+
6
)
2
=
0
Explanation:
Given:
x
3
+
14
x
2
+
60
x
+
72
=
0
By the rational roots theorem, any rational zeros of the given cubic are expressible in the form
p
q
for integers
p
,
q
with
p
a divisor of the constant term
72
and
q
a divisor of the coefficient
1
of the leading term.
That means that the only possible rational zeros are:
±
1
,
±
2
,
±
3
,
±
4
,
±
6
,
±
8
,
±
9
,
±
12
,
±
18
,
±
24
,
±
36
,
±
72
In addition, note that all of the coefficients are positive and the constant term is non-zero. As a result, any real zero (rational or otherwise) of this cubic must be negative.
So that leaves rational possibilities:
−
1
,
−
2
,
−
3
,
−
4
,
−
6
,
−
8
,
−
9
,
−
12
,
−
18
,
−
24
,
−
36
,
−
72
We find:
(
−
2
)
3
+
14
(
−
2
)
2
+
60
(
−
2
)
+
72
=
−
8
+
56
−
120
+
72
=
0
So
x
=
−
2
is a zero and
(
x
+
2
)
a factor:
x
3
+
14
x
2
+
60
+
72
=
(
x
+
2
)
(
x
2
+
12
x
+
36
)
Without trying any more of our "possible" zeros, we can recognise the remaining quadratic factor as a perfect square trinomial:
x
2
+
12
x
+
36
=
x
2
+
2
(
x
)
(
6
)
+
6
2
=
(
x
+
6
)
2
So the factored form of the given cubic equation can be written:
(
x
+
2
)
(
x
+
6
)
2
=
0
The answer is A.) Length = 15, width = 9.
9x2 = 18, 18 - 3 = 15
15 x 2 = 30, 9 x 2 = 18
18 + 30 = 48.
Answer:
They paid 16 3/4% tax
Step-by-step explanation:
9%+1 3/4%+6%=16 3/4%
20: Let 
be the amount of money. If this amount is shared between 8 people, each person will get x/8 dollars. But there actually are 6 people, so everyone is getting x/6 dollars. We know that this difference results in 30 dollars more, so we have
Rearrange the left hand side as
And multiply both sides by 24 to get
21: Let 
and 
be the number of marbles owned by Mike and Judy, respectively. At the beginning, we have
If they both get 9 more marbles, Mike will have 
marbles, and Judy will have 
marbles. So, we have
Plug this value in the first equation and we have
And we deduce
So, the difference is
22: Let 
be the number of $10, $20 and $50 coupon, respectively. We're given:
![\begin{cases}x=2y+3\\z=\frac{1}{2}y\\x+y+z=38\end{cases} We can write the third equation by substituting the expressions for x and z: [tex]x+y+z=(2y+3)+y+\left(\dfrac{1}{2}y\right)=38](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%3D2y%2B3%5C%5Cz%3D%5Cfrac%7B1%7D%7B2%7Dy%5C%5Cx%2By%2Bz%3D38%5Cend%7Bcases%7D%0A%3C%2Fp%3E%3Cp%3EWe%20can%20write%20the%20third%20equation%20by%20substituting%20the%20expressions%20for%20x%20and%20z%3A%0A%3C%2Fp%3E%3Cp%3E%5Btex%5Dx%2By%2Bz%3D%282y%2B3%29%2By%2B%5Cleft%28%5Cdfrac%7B1%7D%7B2%7Dy%5Cright%29%3D38)
Rearrange as follows:
And now we can deduce the number of the other coupons:
So, the total value is
23: a) In order to find the first three terms of the sequence, you just need to plug n=1,2,3:
b) We have
c) 105 is a term of the sequence if and only if there exists an integer k such that
So, 105 is not a term of the sequence.
