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alexira [117]
3 years ago
6

What is the slope of a line perpendicular to the line whose equation is

Mathematics
1 answer:
almond37 [142]3 years ago
6 0

Step-by-step explanation:

4×-20=10y

4x/10-20/10=y

x/5 -2=y

x/5 +(-2)=y--------(1)

we have

y=mx+c----------(2)

comparing eq. 1 and 2,we get,

slope (m)=1/5

c= -2

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Write an equation for a quadratic function that has x intercepts (-3, 0) and (5, 0)
Blizzard [7]

Answer:

One possible equation is f(x) = (x + 3)\, (x - 5), which is equivalent to f(x) = x^{2} - 2\, x - 15.

Step-by-step explanation:

The factor theorem states that if x = x_{0}  (where x_{0} is a constant) is a root of a function, (x - x_{0}) would be a factor of that function.

The question states that (-3,\, 0) and (5,\, 0) are x-intercepts of this function. In other words, x = -3 and x = 5 would both set the value of this quadratic function to 0. Thus, x = -3\! and x = 5\! would be two roots of this function.

By the factor theorem, (x - (-3)) and (x - 5) would be two factors of this function.

Because the function in this question is quadratic, (x - (-3)) and (x - 5) would be the only two factors of this function. In other words, for some constant a (a \ne 0):

f(x) = a\, (x - (-3))\, (x - 5).

Simplify to obtain:

f(x) = a\, (x + 3)\, (x - 5).

Expand this expression to obtain:

f(x) = a\, (x^{2} - 2\, x - 15).

(Quadratic functions are polynomials of degree two. If this function has any factor other than (x - (-3)) and (x - 5), expanding the expression would give a polynomial of degree at least three- not quadratic.)

Every non-zero value of a corresponds to a distinct quadratic function with x-intercepts (-3,\, 0) and (5,\, 0). For example, with a = 1:

f(x) = (x + 3)\, (x - 5), or equivalently,

f(x) = x^{2} - 2\, x - 15.

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Because 300 divided by ten is 30 so 800 divided by ten is 80 :)

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