Option B. one.
A linear equation in two variables is an equation of the form ax + by + c = 0 where a, b, c ∈ R, a, and b ≠ 0. When we consider a system of linear equations, we can find the number of solutions by comparing the coefficients of the variables of the equations.
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{1,3,9}
{3,9}
{3,9,12}
The intersection between the sets are the values that are common to both sets that is {3, 9}
Sets are arrangement of values of elements in a specified way.
Given the following sets
A = {1, 3,5, 7, 9, 11}
B = {3, 6, 9,12}
The intersection between the sets are the values that are common to both sets, hence;
A ∩ B = {3, 9}
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Answer:28
Step-by-step explanation:
Answer:
Step-by-step explanation:
if the coin is tossed four times then the possible sample space is formed
For the probability that the coin is tossed only four times is when
For tossing four times sample space is
S=( HHHH HHHT HHTH HHTT
HTHH HTHT HTTH HTTT
THHH THHT THTH THTT
TTHH TTHT TTTH TTTT )
out of the above required ones are HTHH,TTHH
so the probability is
Answer:
y = 5x + 10
Step-by-step explanation:
using the formula y = mx + b you can easily find the answer.
where
m = the rate of change
b = constant
x = the independent variable
Since he has the initial $10.00 added to the account, we know that is the constant since he won’t be adding that anymore.
Since he will be adding $5.00 to the account every month we know that this is the rate of change.
the variable x will change depending on the number of months passed.
Therefore we can create the equation:
y = 5x + 10
The absolute maximum and minimum of a function on a given interval can be found by calculating the function's critical points and evaluating the function at these points and the interval endpoints, then comparing these values.
In order to find the absolute maximum and absolute minimum values of a function on a given interval, you must first find the critical points of the function within the interval. Critical points occur where the derivative of the function is equal to zero or is undefined. In this case, the derivative of f(t) = 9t + 9 cot(t/2) is f'(t) = 9 - (9/2) csc2(t/2). Set this to zero and solve for t to find the critical points. Additionally, the endpoints of the interval, π/4 and 7π/4, could be the absolute maximum or minimum, so these should be evaluated as well. Once you have found the values of the function at these points and the endpoints, compare them to determine the absolute maximum and minimum values.
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To find the absolute maximum and minimum values of a function, we find the critical points and endpoints. Evaluating the function at these points gives the maximum and minimum values.
To find the absolute maximum and absolute minimum values of a function on a given interval, we need to find the critical points and endpoints of the interval.
To find the critical points of f, we need to find where the derivative of f is equal to zero or undefined. The derivative of f(t) = 9t + 9cot(t/2) is f'(t) = 9 - 9csc^2(t/2).
Setting f'(t) = 0, we have 9 - 9csc^2(t/2) = 0. Solving this equation, we get csc^2(t/2) = 1, which means sin^2(t/2) = 1. This gives us sin(t/2) = ±1. The critical points occur when t/2 = π/2 or t/2 = 3π/2. Solving for t, we get t = π or t = 3π as the critical points.
The endpoints of the interval are π/4 and 7π/4.
Now we evaluate the function f at the critical points and endpoints:
From these evaluations, we can see that the absolute maximum value occurs at t = 7π/4 and is approximately 46.607, while the absolute minimum value occurs at t = π/4 and is approximately 6.566.
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