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22 days ago

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the weight of a piece of wire is directly proportional to its length. A piece of wire is 25cm long and has a weight of 6 grams.

Another piece of the same wire is 30cm long.Calculate the weight of the 30cm piece of wire p is inversely proportional to m.p=48 when m=9 Calculate the value of p when m =12.

1 answer:

Inessa [12.5K]22 days ago

3 0

Part A

The weight of a wire piece correlates directly to its length

⇒ Weight of wire (W) is proportional to Length of wire (L)

⇒ W = k × L

With L = 25 cm, then W = 6 gms

⇒ $\frac{W}{L}$ = k

⇒ k = $\frac{6}{25}$

To find weight for 30 cm wire, substitute k and W into W = k × L

⇒ W = $\frac{6}{25}$ × 30

⇒ W = 7.2 gms

Part B

p is inversely proportional to m

⇒ p = k × m

If p=48 when m=9

k = $\frac{p}{m}$

⇒ k = $\frac{48}{9}$

Now to deduce p when m = 12, replace m and k in p = k × m

p = $\frac{48}{9}$ × 12

⇒ p = 64

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Yes; refer to the question's clarification and the detailed answer below for more information.

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This implies that the question can be interpreted as: can sets A and B be related via a bijective function?

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Two functions are shown in the table below. Function 1 2 3 4 5 6 f(x) = −x2 + 4x + 12 g(x) = −x + 6 Complete the table on your o

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For $\fbox{\begin \\\math{x}=6\\\end{minispace}}$ the function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ both yield the same result.

Detailed breakdown:

The functions involved are

$f(x)=-x^{2}+4x+12$

$g(x)=-x+6$

Step 1:

Insert $x=1$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(1)$ .

$f(1)=-1^{2} +4(1)+12\\f(1)=-1+4+12\\f(1)=15$

Insert $x=1$ in $g(x)=-x+6$ to find the value of $g(1)$ .

$g(1)=-1+6\\g(1)=5$

Step 2:

Insert $x=2$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(2)$ .

$f(2)=-2^{2} +4(2)+12\\f(2)=-4+8+12\\f(2)=16$

Substitute $x=2$ into $g(x)=-x+6$ to find the value of $g(2)$ .

$g(2)=-2+6\\g(2)=4$

Step 3:

Replace $x=3$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(3)$ .

$f(3)=-3^{2} +4(3)+12\\f(3)=-9+12+12\\f(3)=15$

Also, replace $x=3$ in $g(x)=-x+6$ to find the value of $g(3)$ .

$g(3)=-3+6\\g(3)=3$

Step 4:

Insert $x=4$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(4)$ .

$f(4)=-4^{2} +4(4)+12\\f(4)=-16+16+12\\f(4)=12$

Also, replace $x=4$ in $g(x)=-x+6$ to obtain the value of $g(4)$ .

$g(4)=-4+6\\g(4)=2$

Step 5:

Insert $x=5$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(5)$ .

$f(5)=-5^{2} +4(5)+12\\f(5)=-25+20+12\\f(5)=7$

Replace $x=5$ in $g(x)=-x+6$ to find the value of $g(5)$ .

$g(5)=-5+6\\g(5)=1$

Step 6:

Insert $x=6$ into $f(x)=-x^{2} +4x+12$ to find the value of $f(6)$ .

$f(6)=-6^{2} +4(6)+12\\f(6)=-36+24+12\\f(6)=0$

Also, substitute $x=6$ in $g(x)=-x+6$ to obtain the value of $g(6)$ .

$g(6)=-6+6\\g(6)=0$

Step 7:

According to the provided condition $f(x)=g(x)$ .

(a). Insert $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ into the previously mentioned equation.

$-x^{2} +4x+12=-x+6$

(b). Multiply through by $-1$ on both sides.

$x^{2} -4x-12=x-6$

(c). Move the term $x-6$ to the left side of the equation.

$x^{2} -4x-12-x+6=0\\x^{2} -5x-6=0$

(d). Divide the middle term so that its sum equals 5 and the product equals 6.

$x^{2} -(6-1)x-6=0\\x^{2} -6x+x-6=0\\x(x-6)+1(x-6)=0\\(x+1)(x-6)=0\\x=-1,6$

From the analysis above, it is noted that for $x=6$ both functions $f(x)$ and $g(x)$ yield the same outcome.

Using a direct approach:

$f(x)=g(x)\\\Leftrightarrow-x^{2} +4x+12=-x+6\\\Leftrightarrow-x^{2} +4x+12+x-6=0\\\Leftrightarrow-x^{2} +5x+6=0\\\Leftrightarrow-x^{2} +6x-x+6=0\\\Leftrightarrow x^{2} -6x+x-6=0\\\Leftrightarrow x(x-6)+1(x-6)=0\\\Leftrightarrow(x+1)(x-6)=0\\\Leftrightarrow x=6,-1$

The table representing function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ is included below.

For more information:

1. What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0) (2,0)

2. Which is the graph of f(x) = (x – 1)(x + 4)?

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