Ok so if 5 people got an a that means 55 did not receive an a so do 55 over 60 ok they both have a 5 in common 5 goes into 55 11 times and5 goes into 60 12 times so your answer is 11/12
Answer:
agario.work
Step-by-step explanation:
Q1. The answer is <span>D. x4
</span>
Let's first rewrite the expression:
x⁵y²/xy² = x⁵/x * y²/y²
Using the rule xᵃ/xᵇ = x(ᵃ⁻ᵇ), we can write the expression as following:
x⁵y²/xy² = x⁵/x * y²/y² = x⁵⁻¹ * y²⁻² = x⁴ * y⁰ = x⁴ * 1 = x⁴
Thus, the correct answer is D.
Q2. The answer is <span>A. 5(5^3/2/5)^2
</span>
125 in the form of exponent is 5³.
125 = 5³
Now, let's calculate all choices.
The rules we will use are:
xᵃ * xᵇ = x(ᵃ⁺ᵇ)
xᵃ/xᵇ = x(ᵃ⁻ᵇ)
(xᵃ)ᵇ = xᵃ*ᵇ
A. 5(5³/2/5)² = 5 * (5³ * 5/2)²
= 5 * (5³⁺¹/2)²
= 5 * (5⁴/2)²
= 5 * (5⁴)²/(2)²
= 5 * 5⁴*²/4
= 5 * 5⁸ / 4
= 5¹⁺⁸ / 4
= 5⁹/4
≠ 5³ ≠ 125
B. (5³/5⁴)⁻³ = (5³⁻⁴)⁻³
= (5⁻¹)⁻³
= 5⁽⁻¹⁾*⁽⁻³⁾
= 5³
= 125
C. 5⁻²/5⁻⁵ = 5⁽⁻²⁾⁻⁽⁻⁵)
= 5⁽⁻²⁾⁺⁵
= 5³
= 125<span>
D. 5(5</span>⁵/5³) = 5 * 5⁵⁻³
= 5 * 5²
= 5¹⁺²
= 5³
= 125
Therefore, the only expression that is not equal to 125 is A.
Q3. The answer is <span>63x5
Let's check all choices
</span>The rules we will use are:
xᵃ * xᵇ = x(ᵃ⁺ᵇ)
xᵃ/xᵇ = x(ᵃ⁻ᵇ)
(xᵃ)ᵇ = xᵃ*ᵇ
A. 6³<span>x
</span>6³x/6x⁵ = 6³/6 * x/x⁵
= 6³⁻¹ * x¹⁻⁵
= 6²x⁻⁴
= 36x⁻⁴
≠ 36<span>
B. 6</span>³x⁵
6³x⁵/6x⁵ = 6³/6 * x⁵/x⁵
= 6³⁻¹ * x⁵⁻⁵
= 6² * x⁰
<span> = 36 * 1
= 36
C. 6x</span>⁵<span>
</span>6x⁵/6x⁵ = 1
≠ 36
<span>
D. 6</span>⁷x⁵
6⁷x⁵/6x⁵ = 6⁷/6 * x⁵/x⁵
= 6⁷⁻¹ * x⁵⁻⁵
= 6⁶ * x⁰
= 46656 * 1
≠ 36
Therefore, the correct choice is B.
Q4. The answer is
We will use the rule: xᵃ/xᵇ = x(ᵃ⁻ᵇ)
5.4 x 10¹²/1.2 x 10³ = 5.4 / 1.2 x 10¹²/10³
= 4.5 x 10¹²⁻³
= 4.5 x 10⁹
Q5. The answer is B. <span>To subtract powers with the same base, divide the exponents
Some of the rules </span><span>regarding operations with exponents are:
</span>xᵃ/xᵇ = x(ᵃ⁻ᵇ) - choice A
xᵃ * xᵇ = x(ᵃ⁺ᵇ) - choice C
(xᵃ)ᵇ = xᵃ*ᵇ - choice D
Through the process of elimination, choice B is not true.
Answer:
i think its a 4 and half
Step-by-step explanation:
answer
Take the linear coefients easy
11,-12, 7
5,12,-7
so
![\left[\begin{array}{ccc}11&-12&7\\5&12&-7\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%26-12%267%5C%5C5%2612%26-7%5Cend%7Barray%7D%5Cright%5D%20)
or you could do also
![\left[\begin{array}{ccc}11&-12\\5&12\end{array}\right] \left[\begin{array}{ccc}7\\-7\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%26-12%5C%5C5%2612%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%5C%5C-7%5Cend%7Barray%7D%5Cright%5D%20)
