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xenn [34]
1 year ago
15

Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal

to f(x) · g(t)?OA.41 - 81 + 61OB.SIKIAOC.21 + 6I + 2D.6

Mathematics
1 answer:
lisabon 2012 [21]1 year ago
4 0

Given the following functions below,

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}

Factorising the denominators of both functions,

Factorising the denominator of f(x),

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}

Factorising the denominator of g(x),

\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}

Multiplying both functions,

undefined

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Read 2 more answers
Let Q(x, y) be the predicate "If x &lt; y then x2 &lt; y2," with domain for both x and y being R, the set of all real numbers.
Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is Q(x,y) is predicate that x then x^{2}. where x,y are rational numbers.

a)

when x=-2, y=1

Here -2 that is x  satisfied. Then

(-2)^{2}

4 this is wrong. since 4>1

That is x^{2}>y^{2} Thus Q(x,y) =Q(-2,1)is false.

b)

Assume Q(x,y)=Q(-5,2).

That is x=-5, y=2

Here -5 that is x this condition is satisfied.

Then

(-5)^{2}

25 this is not true. since 25>4.

This is similar to the truth value of part (a).

Since in both x satisfied and x^{2} >y^{2} for both the points.

c)

if Q(x,y)=Q(3,8) that is x=3 and y=8

Here 3 this satisfies the condition x.

Then 3^{2}

9 This also satisfies the condition x^{2}.

Hence Q(3,8) exists and it is true.

d)

Assume Q(x,y)=Q(9,10)

Here 9 satisfies the condition x

Then 9^{2}

81 satisfies the condition x^{2}.

Thus, Q(9,10) point exists and it is true. This satisfies the same values as in part (c)

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3 years ago
Divide R300 000 between Leona and Janet in the ratio 3:2.
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Answer:

<em><u>let </u></em><em><u>the </u></em><em><u>ratio</u></em><em><u> be</u></em><em><u> in</u></em><em><u> </u></em><em><u>x </u></em>

<em><u>a/</u></em><em><u>q</u></em><em><u>. </u></em><em><u>3</u></em><em><u>x</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u> </u></em><em><u> </u></em><em><u>5</u></em><em><u>x</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>x </u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>/</u></em><em><u>5</u></em>

<em><u>x=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>Leona </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>*</u></em><em><u>3</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>8</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>Janet </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>*</u></em><em><u>2</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em>

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2 years ago
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