Answer:
(a) Verified
(b) They are simultaneous equations
Step-by-step explanation:
Given
Required
Verify that: 
is a solution
We have:
Substitute:
Evaluate all products
Subtract:
<em>Because both sides of the equation are equal, then the point is a solution</em>
<em></em>
Also:
Substitute:
Evaluate all products
Subtract:
<em>Because both sides of the equation are equal, then the point is a solution</em>
<em></em>
<em></em>
Because the given point is a solution to both equations, then they are simultaneous equation
Answer:
b on edge
Step-by-step explanation:
(x axis 4) y inter -3
49^2m-m : Not equivalent
7^2m-2m : Not equivalent
7^2m-m : Not equivalent
This is actually a trick question. All of the following are actually false statements. Want to know why? Let me show you.
For exponents, if you are dividing a number to some power (i.e 5^3) by the SAME number to a different power (i.e 5^2), then the expression is 5^3-2 or 5^1 = 5. This is true for any number a such that a^b ÷ a^c = a^b-c.
Since 7 and 49 are not the same number, this rule does not apply and thus cannot be simplified any further.
Let me prove why. 5^3 = 125, and 5^2 = 25, and 125 ÷ 25 = 5. This is also the same as 5^3-2 = 5^1 = 5. We just proved this as so.
But, what about 7 and 49, or 2 different numbers. Well it doesn't apply. 7^3 = 343, and 49^2 = 2401, and 343 ÷ 2401 ≈ 0.14. Thus, this is NOT equal to 7^3-2 which is 7. We just proved that a^y ÷ b^z ≠ a^y-z. Congratulations!
Hope this helped!
{1, 2, 4, 5, 7, 7, 8, 9, 9, 10}
Lower Quartile: Q₁ = ¹/₄(n + 1)
Q₁ = ¹/₄(10 + 1)
Q₁ = ¹/₄(11)
Q₁ = 2³/₄
Median: Q₂ = ¹/₂(n + 1)
Q₂ = ¹/₂(10 + 1)
Q₂ = ¹/₂(11)
Q₂ = 5¹/₂
Upper Quartile: Q₃ = ³/₄(n + 1)
Q₃ = ³/₄(10 + 1)
Q₃ = ³/₄(11)
Q₃ = 8¹/₄
Q₁ = 2³/₄
Q₂ = 5¹/₂
Q₃ = 8¹/₄
Answer:
3, 1, - 1, - 3, - 5
Step-by-step explanation:
using the recursive formula and f(1) = 3 , then
f(2) = f(1) - 2 = 3 - 2 = 1
f(3) = f(2) - 2 = 1 - 2 = - 1
f(4) = f(3) - 2 = - 1 - 2 = - 3
f(5) = f(4) - 2 = - 3 - 2 = - 5
first 5 terms are 3, 1, - 1, - 3, - 5
