Answer:
"76°" is the appropriate solution.
Step-by-step explanation:
Please find attachment of the diagram according to the given query.
The given values are:
In ΔDEF,
f = 610 inches
e = 590
∠E = 70°
∠F = ?
By using the law of sines, we get
⇒ 
On substituting the values, we get
⇒ 
On applying cross multiplication, we get
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
now,
⇒ 
⇒ 
Determine the slope of line AB
m = 5
Determine the slope of the lines from the options
First option: y = 5x + 3, the slope is 5
Second option: y = (1/5)x + 3, the slope is 1/5
Third option: y = -5x + 3, the slope is -5
Fourth option: y = (-1/5)x + 3, the slope is -1/5
Parallel lines are similar in the slope. So the line which is parallel to line AB must have the slope of 5.
The answer is first option.
Here are the following justifications:
Given: -2 (x-1) +12=4x-6
Distribute property = -2x+2+12=4x-6
Combine like terms= -2x+14=4x-6
Transpose = -2x = 4x – 20
Addition or Subtraction Property of Equality = -6x = -20
Division property of equality = x = 20/6
Simplify = x = 10/3
Answer:
B. 10 months
Step-by-step explanation:
The balance on the loan will be ...
b = 1600 - 80t . . . . . . where t is the number of months of payments
The balance in the savings account will be ...
s = 500 + 25t
The savings account balance will be at least as much as the loan balance when ...
s ≥ b
500 +25t ≥ 1600 -80t . . . substitute the account balance expressions
105t ≥ 1100 . . . . . . . . . . . . add 80t -500
t ≥ 1100/105 ≈ 10.48 ≈ 10
It will take Josh 10 months to have enough savings to pay the loan in full.
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<em>Comment on rounding</em>
IMO, it makes no sense to round down, as Josh will NOT have enough in 10 months. He will have enough after he makes one more payment of $80. At 10 months, the loan balance is $50 more than the savings balance. It will be 11 months before there is enough savings to pay off the loan.
Answer:
Step-by-step explanation:
The domain of a function is the set for which the function is defined. Our function is the function
. This function is defined regardless of the value of x, so it is defined for every real value of x. That is, it's domain is the set {x|x is a real number}.
The range of the function is the set of all possible values that the function might take, that is {y|y=6x-4}. Recall that every real number y could be written of the form y=6x-4 for a particular x. So the range of the function is the set {y|y is a real number}.
Note that as x gets bigger, the value of 6x-4 gets also bigger, then it doesn't approach any particular number. Note also that as x approaches - infinity, the value of 6x-4 approaches also - infinity. In this case, we don't have any horizontal asymptote. Since the function is defined for every real number, it doesn't have any vertical asymptote. Since h is a linear function, it cannot have any oblique asymptote, then h doesn't have any asymptote.