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Juliette [100K]
3 years ago
10

Please help! It's a grade nine math question.

Mathematics
2 answers:
mash [69]3 years ago
8 0

Answer:

Step-by-step explanation:

Area of the Square

Area = s^2

s = 16

Area = 16 * 16

Area = 256

Area of the Circle.

Area of the Circle = pi*r^2

16 inches is a diameter for the circle.

2r = d

r is the radius

d is the diameter = 16

2r = 16           Divide by 2

2r/2 = 16/2    Do the division

r = 8

Area of the circle = pi*r^2

Area of the circle = 3.14*8^2

Area of the circle = 200.96 cm^2

Area of the shaded area

Area of the shaded area =area of the square - area of the circle

Area of the shaded area = 256 - 200.96

Area of the shaded area = 55.04

 

Jobisdone [24]3 years ago
5 0
Is that all the info they gave you? Idk how to work with it
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3 years ago
What value of b will cause the system to have an infinite number of solutions?
irga5000 [103]

b must be equal to -6  for infinitely many solutions for system of equations y = 6x + b and -3 x+\frac{1}{2} y=-3

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}

Let us bring the equations in same form for sake of simplicity in comparison

\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}

Now we have two equations  

\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}

Let us first see what is requirement for system of equations have an infinite number of solutions

If  a_{1} x+b_{1} y+c_{1}=0 and a_{2} x+b_{2} y+c_{2}=0 are two equation  

\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} then the given system of equation has no infinitely many solutions.

In our case,

\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}

 As for infinitely many solutions \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}

Hence b must be equal to -6 for infinitely many solutions for system of equations y = 6x + b and  -3 x+\frac{1}{2} y=-3

8 0
3 years ago
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