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3 years ago

Find inverse of y= x/2

Mathematics

1 answer:

Anon25 [30]3 years ago

7 0

irst, use the positive value of the

to find the first solution.

√

Next, use the negative value of the

to find the second solution.

−

√

The complete solution is the result of both the positive and negative portions of the solution.

√

−

√

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Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water (in

Bingel [31]

The diver will hit the water when the height is zero, so we will want to set our height function equal to zero
0 = -4t^2 + 11t + 3
we will then factor to find out value of t that make the function equal to zero
0 = (-4t - 1 ) ( t - 3 )
This means our function equals zero when t = -1/4 s and t = 3 seconds
Since time cannot be negative, our final solution is t = 3 seconds

4 0

3 years ago

Read 2 more answers

If alpha and beta are the zeroes of quadratic polynomial ax^2+bx+c , then find alpha^4+beta^4

weqwewe [10]

Ill write A for alpha and B for beta.

AB = c/a and A + B = -b/a

A^4 + B^4 = (A^2 + B^2)^2 - 2A^2B^2

= [(A + B)^2 - 2AB] ^2 - 2A^2B^2

Plugging in the values for A+B and AB we get

A^4 + B^4 = [(-b/a)^2 - 2c/a]^2 - 2(c/a)^2

= (b^2 / a^2 - 2c / a)^2 - 2c^2/a^2

= (b^2 - 2ac)^2 - 2c^2
---------------- -----
a^4 a^2

= (b^2 - 2ac)^2 - 2a^2c^2
-----------------------------
a^4

5 0

3 years ago

Question no 6 answer

ExtremeBDS [4]

Answer:

a = 1
b = -28
P(x) = (x +2)(x -3)(x^2 +3x +10)

Step-by-step explanation:

a) Since x+2 is a factor, we know P(-2) = 0.

P(x) = (((x +2)x +a)x +b)x -60

Then the value of P(-2) is ...

P(-2) = 0 = (((0)(-2) +a)(-2) +b)(-2) -60 = (-2a +b)(-2) -60 = 4a -2b -60

We know the remainder from division by (x+3) is 60, so

P(-3) = 60 = (((-3+2)(-3) +a)(-3) +b)(-3) -60 = ((3+a)(-3) +b)(-3) -60

= (-9 -3a +b)(-3) -60 = 27 +9a -3b -60

93 = 9a -3b

These two equations can be put into standard form:

2a -b = 30

3a -b = 31

Then we have the solution ...

a = 1 . . . . . (by subtracting the first equation from the second)

-28 = b . . . by substituting into the first equation

b) To show that (x-3) is a factor we need to evaluate P(3).

P(3) = (((3 +2)(3) +1)(3) -28)(3) -60 = (48 -28)(3) -60 = 0

The function value is 0, so (x -3) is a factor.

c) We want to find Q(x) = x^2 +cx +d such that ...

(x +2)(x -3)Q(x) = P(x)

(x^2 -x -6)(x^2 +cx +d) = x^4 +2x^3 +x^2 -28x -60

x^4 +(c-1)x^3 +(-6-c+d)x^2 +(-6c-d)x -6d = x^4 +2x^3 +x^2 -28x -60

This gives rise to the equations ...

c -1 = 2 ⇒ c = 3

-6d = -60 ⇒ d = 10

Then P(x) can be factored as ...

P(x) = (x +2)(x -3)(x^2 +3x +10)

_____

<em>Comment on the attached graph</em>

I like to use a graphing calculator to find real roots of higher-degree polynomials. This graph shows the real zeros to be -2 and +3, so we know that (x +2) and (x -3) are factors. The green curve is P(x) with those factors divided out, so is a graph of Q(x). The vertex of that graph tells us that ...

Q(x) = (x +1.5)^2 +7.75 = x^2 +3x +10