In this exact order:
5+1.75x=y
Sorry for the wait..Hope this helps
The given equation has no solution when K is any real number and k>12
We have given that
3x^2−4x+k=0
△=b^2−4ac=k^2−4(3)(12)=k^2−144.
<h3>What is the condition for a solution?</h3>
If Δ=0, it has 1 real solution,
Δ<0 it has no real solution,
Δ>0 it has 2 real solutions.
We get,
Δ=k^2−144 here Δ is not zero.
It is either >0 or <0
Δ<0 it has no real solution,
Therefore the given equation has no solution when K is any real number.
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brainly.com/question/1397278
Answer:
No
Step-by-step explanation:
Let's calculate sinA and sinB.
sinA = opposite / hypotenuse = 12 / 20
sinB = opposite / hypotenuse = 16 / 20
Since 12 / 20 ≠ 16 / 20, she is incorrect.
Answer:
5.525
Step-by-step explanation:
1. (7x^(2))/(2x + 6) -:(3x - 5) / (x + 3) Solve Bold
(7x^(2)) / (2x + 6) -:(3x - 5) / (x + 3)
2. 49x / 8x - (-2x) / 3x Solve Bold
49x/8x-(-2x)/3x
3. 6.125 - .6 Solve Bold
Answer: 5.525
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