Answer:
None
Step-by-step explanation:
We are given a triangle ABC with ∠A = 30°, sides a = 4 and b = 10.
According to the 'Law of Sines- Ambiguous Case', we have,
If a < b×sinA, then no triangle is possible.
If a = b×sinA, only one triangle is possible
If a > b×sinA, two triangles are possible.
So, we have,
b×sinA = 10 × sin30 = 10 × 0.5 = 5.
Now, as
4 = a < bsinA = 5.
We get, according to the rule, no triangle is possible.
Answer:
60,000
Step-by-step explanation:
2813 rounds to 60,000 not 61,000
You would use this formula for the problem...
(C=2πr) The circumference equals 2 multiplied by pi and radius.
In this problem it says diameter, but the diameter is just twice the length of the radius, so just multiply the radius by 2. Plugin your number.
C=2π(2) since 2 is half of 4, that's your radius.
Your answer is roughly C= 12.57.
You're welcome.
For the answer to the question above, For θ the range is between -infinity and infinity,
<span>While y falls in the range from -1 to 1.
because </span><span>y is an element of [-1, 1]</span>
I hope my answer helped you in your problem. Have a nice day!
The vertex form of the equation f(x) = x^2 - 3x, is f(x) = (x - 3/2)^2 - 9/5
<h3>How to rewrite the
quadratic function?</h3>
The quadratic function is given as:
f(x) = x^2 - 3x
Differentiate the function
f'(x) = 2x - 3
Set the function to 0
2x - 3 = 0
Add 3 to both sides
2x = 3
Divide by 2
x = 3/2
Set x = 3/2 in f(x) = x^2 - 3x
f(x) = 3/2^2 - 3 * 3/2
Evaluate
f(x) = -9/5
So, we have:
(x, f(x)) = (3/2, -9/5)
The above represents the vertex of the quadratic function.
This is properly written as:
(h, k) = (3/2, -9/5)
The vertex form of a quadratic function is
f(x) = a(x - h)^2 + k
So, we have:
f(x) = a(x - 3/2)^2 - 9/5
In f(x) = x^2 - 3x,
a = 1
So, we have:
f(x) = (x - 3/2)^2 - 9/5
Hence, the vertex form of the equation f(x) = x^2 - 3x, is f(x) = (x - 3/2)^2 - 9/5
Read more about vertex form at
brainly.com/question/24850937
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