Hey!!
here is your answer >>>
1) 3x - 10 > 11
let's take this as 3x - 10 = 11
3x = 21
x = 7
So, The numbers of X are always more than 7
2) 9x + 2 < 5
9x + 2 = 5
9x = 3
x = 1÷2
The numbers of X are always less than 1 ÷ 2
3) 2x + 8 < 6
2x + 8 = 6
2x = 2
x = 1
The numbers of X are less than 1
4) 4x - 3 < 12
4x - 3 = 12
4x = 15
x = 15 ÷ 4
The numbers of X are less than 15 ÷ 4
5) 2 ( 9 + x ) > 20
2 ( 9 + x ) = 20
9 + x = 10
x = 1
The numbers of X are more than 1
We can solve the rest using this
Hope my answer helps!
(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.
We can factor this by using grouping. Take the leading coefficient and multiply it by the constant. In this case we get 5*-7 = -35.
Now we need 2 numbers that add to 2 and multiply to be -35. The numbers are -5 and 7.
So split the 2r into these two terms and group.
(5r^2 - 5r) + (7r - 7)
Factor both groups.
5r(r-1) + 7(r-1)
The factors of (r-1) can be added together to get the answer.
(r-1)(5r+7)
Hi there,
Sol:
Length of the swimming pool = 260 m
Breadth of the swimming pool = 140 m
Let the height of the water in the swimming pool after pouring water be h meters.
Volume of the water poured = 54600 m3
l x b x h = 54600
260 x 140 x h = 54600
h = (54600) / (260 x 140)
h = 1.5 m = 150 cm
Hope it helps
Answer: the area of the triangle is 23.4 cm²
Step-by-step explanation:
The given triangle is not a right angle triangle. Since two sides and one angle are known, we can either apply the Heron's formula or the Sine formula which is expressed as
Area of triangle = 1/2abSinC
Where a and b are the sides of the triangle and C is the given angle. Therefore,
Area = 1/2 × 8.2 × 6.4 × Sin63
Area = 1/2 × 8.2 × 6.4 × 0.8910
Area = 23.4 cm² to the nearest tenth.
