Solve the literal equation for y. 2x – xy = 3
2 answers:
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Answer:
if x is odd then x> 0 is true statement.
if x is odd then x> 0 is true statement.
if x is even then x≤0 is false statement.
If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.
if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.
Step-by-step explanation:
Consider the provided information.
D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36}
Part (A) if x is odd then x> 0
Here only even numbers are less than 0 that means the statement is true.
if x is odd then x> 0 is true statement.
Part (B) if x is less than 0 then x is even.
Here only even numbers are less than 0 that means the statement is true.
if x is odd then x> 0 is true statement.
Part (C) if x is even then x≤0
Here we can see that 16, 26, 32, 36 are even number and also greater than 0. Thus the statement is false.
if x is even then x≤0 is false statement.
Part (D) If the ones digit of x is 2, then the tens digit is 3 or 4.
There is only one number whose ones digit is 2. i.e. 32 also the tens digit of the number 32 is 3. Which makes the above statement true.
If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.
Part (E) if the ones digit of x is 6, then the tens digit is 1 or 2.
Numbers having ones digit 6 are: 16, 26 and 36
Here, the tens digits are 1, 2 and 3 which is contradict to our statement. Hence the provided statement is false.
if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.
Answer:
and
Step-by-step explanation:
Assume that the terminal side of thetaθ passes through the point (−12,5).
In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.
Using Pythagoras theorem:
Taking square root on both sides.
In a right angled triangle
In second quadrant only sine and cosecant are positive.
and
