we have
x=5−2y ---> multiply by 3 both sides-----> 3x=15-6y-----> 6y=15-3x----> equation 1
6y=10−3x------> equation 2
equation 1 and equation 2 has the same slope
the lines are parallel
slope m=-3/6-----> m=-1/2
so
there are no solution
the answer is the option
b. There is no solution
Answer:
To convert this fraction to a decimal, just divide the numerator (5) by the denominator (8): 5 ÷ 8 = 0.625. So,
5/8 = 0.625
Rounded values:
5/8 = 1 rounded to the nearest integer5/8 = 0.6 rounded to 1 decimal place5/8 = 0.63 rounded to 2 decimal places
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Step-by-step explanation:
Part A: This data does represent a function because no value of x is repeated.
Part B: If x = 6 for the relation f(x) = 7x - 15, then f(x) will equal 27. The function f(x) = 7x - 15 will have the greater value
Part C: If f(x) is equal to 6 then x will equal 3
1. We use the recursive formula to make the table of values:
f(1) = 35
f(2) = f(1) + f(2-1) = f(1) + f(1) = 35 + 35 = 70
f(3) = f(1) + f(3-1) = f(1) + f(2) = 35 + 70 = 105
f(4) = f(1) + f(4-1) = f(1) + f(3) = 35 + 105 = 140
f(5) = f(1) + f(5-1) = f(1) + f(4) = 35 + 140 = 175
2. We observe that the pattern is that for each increase of n by 1, the value of f(n) increases by 35. The explicit equation would be that f(n) = 35n. This fits with the description that Bill saves up $35 each week, thus meaning that he adds $35 to the previous week's value.
3. Therefore, the value of f(40) = 35*40 = 1400. This is easier than having to calculate each value from f(1) up to f(39) individually. The answer of 1400 means that Bill will have saved up $1400 after 40 weeks.
4. For the sequence of 5, 6, 8, 11, 15, 20, 26, 33, 41...
The first-order differences between each pair of terms is: 1, 2, 3, 4, 5, 6, 7, 8...since these differences form a linear equation, this sequence can be expressed as a quadratic equation. Since quadratics are functions (they do not have repeating values of the x-coordinate), therefore, this sequence can also be considered a function.
ANSWER
The solution is 
EXPLANATION
We want to solve the simultaneous equations
and
.
We substitute equation (2) in to equation (1), to obtain
This has now become a linear equation in a single variable 
.
We solve for x by grouping like terms.
We divide through by negative 3 to get;
.
Hence, the solution is
