Answer:
The two positive integers are 30 and 30
Step-by-step explanation:
Let
x------> the larger positive integer
y-----> the smaller positive integer
P----> the product o the two positive integers
we know that
-----> equation A
----> equation B
substitute equation A in equation B
This is the equation of a vertical parabola open downward
The vertex is a maximum
The y-coordinate of the vertex is the largest possible product
Using a graphing tool
The vertex is the point (30,900)
That means------> For 
The largest possible product is 
and
------> 
therefore
The two positive integers are 30 and 30
Answer:
value of x = 4
Step-by-step explanation:
here's the solution : -
=》6 - 2x = 5x - 9x + 8
shifting like terms aside,
=》6 - 8 = 5x - 9x + 2x
\_(●_●)_/
=》-2 = -2x
=》x = -2 / -2
=》x = 1
(a) ![[\frac{9}{2.6} - \frac{2.5^{2} }{2.5} ]^{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B9%7D%7B2.6%7D%20%20-%20%5Cfrac%7B2.5%5E%7B2%7D%20%7D%7B2.5%7D%20%5D%5E%7B2%7D)
Answer:
= ![[\frac{9}{2.6} - \frac{2.5*2.5 }{2.5} ]^{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B9%7D%7B2.6%7D%20%20-%20%5Cfrac%7B2.5%2A2.5%20%7D%7B2.5%7D%20%5D%5E%7B2%7D)
= ![[\frac{9}{2.6} - \frac{2.5}{1} ]^{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B9%7D%7B2.6%7D%20%20-%20%5Cfrac%7B2.5%7D%7B1%7D%20%5D%5E%7B2%7D)
*canceling 2.5 in numerator and denominator*
![= [\frac{9-(2.5)(2.6)}{2.6} ]^2\\*Using L.C.M of 2.6 and 1 which comes out to be '2.6'= [\frac{9-(6.5)}{2.6} ]^2\\= [\frac{2.5}{2.6} ]^2\\*multiplying and dividing by '10'= [\frac{2.5*10}{2.6*10} ]^2\\= [\frac{25}{26} ]^2\\= \frac{25^2}{26^2}\\= \frac{625}{676}\\= 0.925](https://tex.z-dn.net/?f=%3D%20%5B%5Cfrac%7B9-%282.5%29%282.6%29%7D%7B2.6%7D%20%5D%5E2%5C%5C%3C%2Fp%3E%3Cp%3E%2AUsing%20L.C.M%20of%202.6%20and%201%20which%20comes%20out%20to%20be%20%272.6%27%3C%2Fp%3E%3Cp%3E%3D%20%5B%5Cfrac%7B9-%286.5%29%7D%7B2.6%7D%20%5D%5E2%5C%5C%3D%20%5B%5Cfrac%7B2.5%7D%7B2.6%7D%20%5D%5E2%5C%5C%3C%2Fp%3E%3Cp%3E%2Amultiplying%20and%20dividing%20by%20%2710%27%3C%2Fp%3E%3Cp%3E%3D%20%5B%5Cfrac%7B2.5%2A10%7D%7B2.6%2A10%7D%20%5D%5E2%5C%5C%3D%20%5B%5Cfrac%7B25%7D%7B26%7D%20%5D%5E2%5C%5C%3D%20%5Cfrac%7B25%5E2%7D%7B26%5E2%7D%5C%5C%3D%20%5Cfrac%7B625%7D%7B676%7D%5C%5C%3D%200.925)
Properties used:
Cancellation property of fractions
Least Common Multiplier(LCM)
The least or smallest common multiple of any two or more given natural numbers are termed as LCM. For example, LCM of 10, 15, and 20 is 60.
(b) ![[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3} ] ^{2}](https://tex.z-dn.net/?f=%20%5B%5B%5Cfrac%7B3x%5E%7Ba%7Dy%5E%7Bb%7D%7D%20%7B-3x%5E%7Ba%7D%20y%5E%7Bb%7D%20%7D%20%5D%5E%7B3%7D%20%20%20%20%5D%20%5E%7B2%7D%20)
Answer:
![[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3}] ^{2}\\](https://tex.z-dn.net/?f=%5B%5B%5Cfrac%7B3x%5E%7Ba%7Dy%5E%7Bb%7D%7D%20%7B-3x%5E%7Ba%7D%20y%5E%7Bb%7D%20%7D%20%5D%5E%7B3%7D%5D%20%5E%7B2%7D%5C%5C)
*using ![[x^{a}]^b = x^{ab}](https://tex.z-dn.net/?f=%5Bx%5E%7Ba%7D%5D%5Eb%20%3D%20x%5E%7Bab%7D)
*
![= [\frac{3x^{3a}y^{3b}} {-3x^{3a} y^{3b} }] ^{2}](https://tex.z-dn.net/?f=%3D%20%5B%5Cfrac%7B3x%5E%7B3a%7Dy%5E%7B3b%7D%7D%20%7B-3x%5E%7B3a%7D%20y%5E%7B3b%7D%20%7D%5D%20%5E%7B2%7D)
*Again, using *
![= \frac{3x^{2*3a}y^{2*3b}} {-3x^{2*3a} y^{2*3b} } \\= (-1)\frac{3x^{6a}y^{6b}} {3x^{6a} y^{6b} }\\[\tex]*taking -1 common, denominator and numerator are equal*[tex]= -(1)\frac{1}{1}\\= -1](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B3x%5E%7B2%2A3a%7Dy%5E%7B2%2A3b%7D%7D%20%7B-3x%5E%7B2%2A3a%7D%20y%5E%7B2%2A3b%7D%20%7D%20%20%5C%5C%3D%20%28-1%29%5Cfrac%7B3x%5E%7B6a%7Dy%5E%7B6b%7D%7D%20%7B3x%5E%7B6a%7D%20y%5E%7B6b%7D%20%7D%5C%5C%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3E%2Ataking%20-1%20common%2C%20denominator%20and%20numerator%20are%20equal%2A%3C%2Fp%3E%3Cp%3E%5Btex%5D%3D%20-%281%29%5Cfrac%7B1%7D%7B1%7D%5C%5C%3D%20-1)
Property used: 'Power of a power'
We can raise a power to a power
(x^2)4=(x⋅x)⋅(x⋅x)⋅(x⋅x)⋅(x⋅x)=x^8
This is called the power of a power property and says that to find a power of a power you just have to multiply the exponents.
Answer:
Step-by-step explanation:
x
2
+
x
−
6
=
(
x
+
3
)
(
x
−
2
)
x
2
−
3
x
−
4
=
(
x
−
4
)
(
x
+
1
)
Each of the linear factors occurs precisely once, so the sign of the given rational expression will change at each of the points where one of the linear factors is zero. That is at:
x
=
−
3
,
−
1
,
2
,
4
Note that when
x
is large, the
x
2
terms will dominate the values of the numerator and denominator, making both positive.
Hence the sign of the value of the rational expression in each of the intervals
(
−
∞
,
−
3
)
,
(
−
3
,
−
1
)
,
(
−
1
,
2
)
,
(
2
,
4
)
and
(
4
,
∞
)
follows the pattern
+
−
+
−
+
. Hence the intervals
(
−
3
,
−
1
)
and
(
2
,
4
)
are both part of the solution set.
When
x
=
−
1
or
x
=
4
, the denominator is zero so the rational expression is undefined. Since the numerator is non-zero at those values, the function will have vertical asymptotes at those points (and not satisfy the inequality).
When
x
=
−
3
or
x
=
2
, the numerator is zero and the denominator is non-zero. So the function will be zero and satisfy the inequality at those points.
Hence the solution is:
x
∈
[
−
3
,
−
1
)
∪
[
2
,
4
)
graph{(x^2+x-6)/(x^2-3x-4) [-10, 10, -5, 5]}
