Answer:
16/25
Step-by-step explanation:
got it on calculator
Answer:
width = 150 yards
length = 200 yards
Step-by-step explanation:
Using the information given to us, we can make the following two equations:
2L + 2W = 700
L = W + 50
where L is the length in yards and W is the width in yards. This could treat these two equations as a system of linear equations. There are multiple ways to solve a system of linear equations, but I am going to solve this by substitution. Before doing that, I am first going to simplify the first equation, as it will make solving the system easier.
2L + 2W = 700
Divide both sides by 2.
L + W = 350
Since L = W + 50, we can substitute W + 50 for L in the equation L + W = 700.
(W + 50) + W = 350
2W + 50 = 350
2W = 300
W = 150
Now that we found what W is, we can solve for L by plugging 150 into either one of the equations. I am going to plug it into the equation L = W + 50.
L = 150 + 50
L = 200
Now we found what L is. So now we know that the width is 150 yards and the length is 200 yards.
I hope you find my answer and explanation helpful. Happy studying. :)
Hi there!
h(x) = 5x + 11
To find h(7) - 3, we can substitute 7 for x and subtract 3:
h(7) = 5(7) + 11 - 3
Simplify:
h(7) = 35 + 11 - 3
h(7) = 43
De l'Hospital rule applies to undetermined forms like
If we evaluate your limit directly, we have
which is neither of the two forms covered by the theorem.
So, in order to apply it, we need to write the limit as follows: we start with
Using the identity , we can rewrite the function as
Using the rule , we have
Since the exponential function is continuous, we have
In other words, we can focus on the exponent alone to solve the limit. So, we're focusing on
Which we can rewrite as
Now the limit comes in the form 0/0, so we can apply the theorem: we derive both numerator and denominator to get
So, the limit of the exponent is -6, which implies that the whole expression tends to
Answer:
The three quadratic equations are;
x² + 3 = 0
x² + 2·x + 1 = 0
x² + 3·x + 2 = 0
Step-by-step explanation:
1) A quadratic equation with no real solution is one with an imaginary solution such as one with a negative square root
We can write the quadratic equation as follows;
x² + 3 = 0
∴ x = √(-3) = √(-1) ×√3 = i·√(3)
Therefore, the equation f(x) = x² + 3, has no real root at f(x) = 0
2) A quadratic that has 1 real root is of the form;
(x + 1)² = 0
The root of the equation is x = -1 from (x + 1) = ((-1) + 1)² = 0²
Which gives;
(x + 1)² = (x + 1)·(x + 1) = x² + 2·x + 1 = 0
Therefore, the quadratic (x + 1)² = 0 has only one real root
3) A quadratic that has 2 real root is of the form;
(x + 1)·(x + 2) = 0
x² + x + 2·x + 2 = 0
x² + 3·x + 2 = 0
Therefore, the three quadratic equations are;
x² + 3 = 0
x² + 2·x + 1 = 0
x² + 3·x + 2 = 0