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 #3
avatar+130536 
0
Sep 9, 2019
 #2
avatar+234 
+1
Sep 9, 2019
 #2
avatar+26400 
+2

There are 360 people in my school.

15 take calculus, physics, and chemistry, and

15 don't take any of them.

180 take calculus.

Twice as many students take chemistry as take physics.

75 take both calculus and chemistry, and

75 take both physics and chemistry.

Only 30 take both physics and calculus.

 

How many students take physics? 

 

\(\begin{array}{|c|c|c|rl|} \hline \text{calculus} & \text{physics} & \text{chemistry} & \\ \hline x & x & x & 15 & \text{ take calculus, physics, and chemistry} \\ \hline x & & x & 75 & \text{ take both calculus and chemistry} \\ \hline x & x & & 30 & \text{ take both physics and calculus} \\ \hline x & & & 180-15-75-30 = \mathbf{60} & \text{ take only calculus} \\ \hline & & & 180 & \text{ take calculus} \\ \hline \end{array} \)

 

\(\begin{array}{|c|c|c|rl|} \hline \text{chemistry} & \text{physics} & \text{calculus} & \\ \hline x & x & x & 15 & \text{ take calculus, physics, and chemistry} \\ \hline x & & x & 75 & \text{ take both calculus and chemistry} \\ \hline x & x & & 75 & \text{ take both physics and chemistry} \\ \hline x & & & v & \text{ take only chemistry} \\ \hline & & & 15+75+75+v \\ & & & =165+v & \text{ take chemistry} \\ \hline \end{array}\)

 

\(\begin{array}{|c|c|c|rl|} \hline \text{physics} & \text{chemistry} & \text{calculus} & \\ \hline x & x & x & 15 & \text{ take calculus, physics, and chemistry} \\ \hline x & & x & 30 & \text{ take both physics and calculus} \\ \hline x & x & & 75 & \text{ take both physics and chemistry} \\ \hline x & & & u & \text{ take only physics} \\ \hline & & & 15+30+75+u \\ & & & =120+u & \text{ take physics} \\ \hline \end{array}\)

 

Twice as many students take chemistry as take physics:

\(\begin{array}{|rcll|} \hline 165+v &=& 2(120+u) \\ &=& 240 +2u \\ \mathbf{v} &=& \mathbf{2u+75} \\ \hline \end{array}\)

 

There are 360 people in my school:

\(\begin{array}{|rcll|} \hline \overbrace{180}^{\text{take calculus}} + \overbrace{u+v+ \underbrace{75}_{\text{take both physics and chemistry} } + \underbrace{15}_{\text{don't take any of them}} }^{\text{take no calculus}} &=& 360 \\ 180 + u+v+ 75 +15 &=& 360 \\ u+v+ 90 &=& 180 \\ u+v &=& 90 \\ \mathbf{u} &=& \mathbf{90-v} \quad | \quad v=2u+75 \\ u &=& 90-(2u+75) \\ u &=& 90-2u-75 \\ 3u &=& 15 \\ \mathbf{u} &=& \mathbf{5} \\\\ v &=& 2u+75 \\ v &=& 2*5 + 75 \\ \mathbf{v} &=& \mathbf{85} \\ \hline \end{array} \)

 

How many students take physics?

\(\begin{array}{|rcll|} \hline && 120+u \text{ take physics} \\ &=& 120 +5 \\ &=& 125 \text{ take physics} \\ \hline \end{array} \)

 

laugh

Sep 9, 2019
 #1
avatar
+1
Sep 9, 2019
 #1
avatar+118724 
+2

g(x) is a concave up parabola

 

\(g(x)=x^2-8x+7\\ g(x)=(x-7)(x-1)\\ \text{roots are at x= 7 and x= 1}\\ \text{Axis of symmetry = 4} \)

 

 x=4    g(4)= -3*3 = -9

So the vertex is (4,-9)

Since the domain is \((4,\infty)\)     the range is  \((-9,\infty) \)

 

So the domain and range of the inverse must be the other way around.

 

For \(g^{-1}(x)\)

The domain is \((-9,\infty) \)  abd the range is \((4,\infty)\)

 

Here is the graph.

https://www.desmos.com/calculator/lgob2ihohq

 

 

 

 

 

This maths underneath is not needed to answer the question but I have done it so I might as well leave it.

 

 

 

\(Let\;\;y=g(x)\\ y=x^2-8x+7 \qquad x>4\\ y-7=x^2-8x\\ y-7+16=x^2-8x+16\\ y+9=(x-4)^2\\ \pm\sqrt{y+9}=x-4\\ \text{But x-4 is positive, so}\\ x-4=\sqrt{y+9}\\ x=\sqrt{y+9}+4\\ g^{-1}(x)=\sqrt{x+9}+4\)

.
Sep 9, 2019
 #3
avatar+2864 
+3

or you can use this website's 

 

wink

Sep 9, 2019
Sep 8, 2019

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