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Ymorist [56]
3 years ago
13

consider the point of intersection where the vertical line x=2 meets the line y=7x+1 what is the y-coordinate​

Mathematics
1 answer:
VMariaS [17]3 years ago
5 0
<h3>Answer:  15</h3>

=======================================================

Work Shown:

Replace every x with 2 and simplify

y = 7x+1

y = 7*2 + 1

y = 14 + 1

y = 15 is the y coordinate

The vertical line x = 2 meets the diagonal line y = 7x+1 at the location (x,y) = (1,15)

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Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
likoan [24]

Answer:

Thus, the two root of the given quadratic equation x^2+4=6x is 5.24 and 0.76 .

Step-by-step explanation:

Consider, the given Quadratic equation, x^2+4=6x

This can be written as ,  x^2-6x+4=0

We have to solve using quadratic formula,

For a given quadratic equation ax^2+bx+c=0 we can find roots using,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}  ...........(1)

Where,  \sqrt{b^2-4ac} is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}

\Rightarrow x=\frac{6\pm\sqrt{20}}{2}

\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}

\Rightarrow x={3\pm \sqrt{5}}

\Rightarrow x_1={3+\sqrt{5}} and \Rightarrow x_2={3-\sqrt{5}}

We know \sqrt{5}=2.23607(approx)

Substitute, we get,

\Rightarrow x_1={3+2.23607}(approx) and \Rightarrow x_2={3-2.23607}(approx)

\Rightarrow x_1={5.23607}(approx) and \Rightarrow x_2=0.76393}(approx)

Thus, the two root of the given quadratic equation x^2+4=6x is 5.24 and 0.76 .

7 0
3 years ago
Read 2 more answers
Can someone explain this please???
xeze [42]

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3. Making the nurse pool the domain and the patient pool the range would give a relation that is not a function, since more than one patient is assigned to one nurse.

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vova2212 [387]

Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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