the first question...the answer is 10
see;x is our unknown number
so;27-2=25
x + 15=25
x = 25-15
x =10
10 + 15 = 25
25 + 2 = 27
Given that the equation is y = 37 + 5, to find the coordinate of (2, ?), we substitute the x coordinate, x = 2, into the equation.
y = 37x + 5
Substitute x = 2,
y = 37 (2) + 5
y = 79
So now that we know that the y is 79, we know that coordinate is (2, 79).
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Answer: The other half of the coordinate is 79.
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In general to be able to add or subtract fractions you need to have the same denominator so that you can add the numerators.
The easiest way to find the same denominator is to identify the lowest common denominator (LCD).
You can find the LCD by finding the prime factors of the denominators in question and multiplying them all together. If the denominators share a prime factor, only multiply it once.
Sometimes you can just eyeball the numbers to find the LCD, which might be faster.
For #W we need to find the LCD of 10 and 6, so prime factorize:
10 = 2 x 5
6 = 2 x 3
LCD = 2 x 5 x 3 = 30
The LCD is 30, so we need to change the fraction to reflect that. Remember, what you do to the denominator you need to do to the numerator as well. So:
-9/10 becomes -27/30 (both multiplied by 3)
-1/6 becomes -5/30 (both multiples by 5)
Now you can easily add:
-27/30 + (-5/30) = -32/30
In summary:
Step #1: find LCD (prime factor or eyeball)
Step #2: multiply the numerator of each fraction by the factor needed to obtain the LCD in that denominator
Step #3: add the fractions now that they have the LCD
Here’s the solution to #G:
-¾ + (-7/12)
Step #1:
4 = 2 x 2
12 = 2 x 2 x 3
LCD = 2 x 2 x 3 (count each unique prime factor once)
Step #2:
-¾ becomes -9/12 (both multiplied by 3)
-7/12 stays the same (it already has the LCD)
Step #3:
-9/12 + (-7/12) = -16/12
Let me know if you have any questions. Try to work though the others!
Answer:
7
Step-by-step explanation:
:)
i) The given function is
The factored form is
The domain are the values of x for which the function is defined.
ii) To find the vertical asymptotes, equate the denominator to zero.
iii) To find the roots, equate the numerator to zero.
The root is
iv) To find the y-intercept, put into the function.
The y-intercept is
v) The horizontal asymptote is given by;
The horizontal asymptote is
vi) The function is not reducible. There are no holes.
vii) The given function is a proper rational function.
Proper rational functions do not have oblique asymptotes.