ANSWER
2x-3=0
and
x+4=0
EXPLANATION
Given that:
Then to find roots of
is the same as finding the roots of
By the zero product property, we must have either,
Hence the equations that must be solved to find the roots are:
2x-3=0
and
x+4=0
Answer:
264
Step-by-step explanation:
Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
Start with the last two choices.
You have x + 5 and x - 5 in the square root.
A square root can never be negative.
That means that the lowest value y can have in the last two choices is 0.
The range does not include -4.
Now look at choice B.
sqrt(x) is at least 0.
Then you add 5.
The range does not include -4.
Now look at choice A.
sqrt(x) is at least 0.
Now you subtract 5.
The lowest value y can have is -5. That includes -4.
Choice A is the answer.
y will be at least 5.
Answer:
The correct options are;
Adding rational expressions
Subtracting rational expressions
Step-by-step explanation:
When adding or subtracting rational numbers, the denominators of the fractions representing the numbers should be equal
Therefore, when adding two or more rational expressions together, the following steps are required;
1) Looking for the Lowest Common Denominator
2) Finding the product of the quotient obtained from dividing the LCD by the denominator of a fraction and the numerator of the fraction
3) Placing the sum (or differences) of the products obtained in the above step as the numerator, while the LCD remain denominator
4) Simplifying the numerator in the above step
5) The faction so obtained, is the sum or difference of two or more rational expressions.