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3 years ago

How many solutions will the system of equations have? System: y = 2x + 2 and y = x + 4

1 answer:

6 0

Y=-2x+2
y=x2-2x+4

is the value of x such that when you plug it into both equations, they both produce the same value of y, or the value of x such that

-2x+2=x2-2x+4

solving this equation----------> x2=-2, taking square root of both sides gives x= ±√2i, or the system has no real solutions but has 2 complex solutions. If you graph both functions in the xy-plane, where x, y are real numbers, you' ll see they do not intersect, which implies there is no real number solutions.

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Solve for m . − 5 m = − 35 m =

dolphi86 [110]

Answer: 7

Step-by-step explanation: -35 / -5 = 7

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3 years ago

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Find the length of the third side. If necessary, round to the nearest tenth. 15 9

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3 years ago

If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

$= \dfrac{-1}{x(x+h)}, h\ne 0$

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3 years ago

The curve with equation y = x - 3x + 1 is reflected in the x-axis.

ankoles [38]

Answer:

Step-by-step explanation:

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3 years ago

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Sliva [168]

11^2 = h^2 + ( 14/2)^2

11^2 = h^2 + 7^2

121 = h^2 + 49

h^2 = 121 - 49

h^2 = 72

h = √72

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3 years ago