Answer:
the first number is 5 the second number is 4
Step-by-step explanation:
first number = x
second number = y
X+2y=13 given this isolate x x = 13-2y
2x+y=14 substitute 13-2y for x
2(13-2y)+y=14
26-4y+y=14
-3y=-12
y=4 taking this number substitute into either equation above
x+2(4)=13
x+8=13
x=5
Answer:
Step-by-step explanation:
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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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(C)
Step-by-step explanation:
The volume of the conical pile is given by

Taking the derivative of V with respect to time, we get


Since r is always equal to h, we can set

so that our expression for dV/dt becomes


Solving for dh/dt, we get


