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earnstyle [38]
3 years ago
5

Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than

or equal to y. Also, the sum of y and two-third of x is less than 4. Which graph represents the possible solutions?

Mathematics
1 answer:
Fittoniya [83]3 years ago
4 0
Inequation 1: 

\frac{2}{3}x-3 \geq y

to plot the pairs (x, y) for which the inequation holds, draw the line y=\frac{2}{3}x-3

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2: 

y+ \frac{2}{3}x\ \textless \ 4

y\ \textless \ - \frac{2}{3} x+ 4



draw the lines 

i)  y=\frac{2}{3}x-3          use points (0, -3),  (3, -1)

ii)y=- \frac{2}{3} x+ 4       use points ( 0, 4),   (3, 2)


let's use the point P(3, 3) to see what region of the lines need to be coloured:

\frac{2}{3}x-3 \geq y  ; 
\frac{2}{3}(3)-3 \geq 3
2-3 \geq 3, not true so we color the region not containing this point


y+ \frac{2}{3}x\ \textless \ 4
(3)+ \frac{2}{3}(3)\ \textless \ 4
3+ 5\ \textless \ 4 not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.



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Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the trian
777dan777 [17]

Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

<em>First case</em>

In the triangle ABC

Let

a=4\ units\\b=2/ units\\B=30^o

Applying the law of sines

Find the measure of angle A

\frac{a}{sin(A)}=\frac{b}{sin(B)}

substitute the given values

\frac{4}{sin(A)}=\frac{2}{sin(30^o)}

sin(A)=1

so

A=90^o

Find the measure of angle C

In a right triangle

we know that

B+C=90^o ----> by complementary angles

B=30^o

therefore

C=60^o

Find the length side c

Applying the law of sines

\frac{c}{sin(C)}=\frac{b}{sin(B)}

substitute the given values

\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}

c=2\sqrt{3}\ units

therefore

The dimensions of the triangle are

A=90^o

B=30^o

C=60^o

a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units

<em>Second case</em>

In the triangle ABC

Let

a=4\ units\\b=2/ units\\A=30^o

Applying the law of sines

Find the measure of angle B

\frac{a}{sin(A)}=\frac{b}{sin(B)}

substitute the given values

\frac{4}{sin(30^o)}=\frac{2}{sin(B)}

sin(B)=0.25

so

using a calculator

B=14.48^o

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

A+B+C=180^o

A=30^o\\B=14.48^o

therefore

30^o+14.48^o+C=180^o

C=135.52^o

Find the length side c

Applying the law of sines

\frac{c}{sin(C)}=\frac{a}{sin(A)}

substitute the given values

\frac{c}{sin(135.52^o)}=\frac{4}{sin(30^o)}

c=5.61\ units

therefore

The dimensions of the triangle are

A=30^o

B=14.48^o

C=135.52^o

a=4\ units\\b=2\ units\\c=5.61\ units

see the attached figure to better understand the problem

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Answer:

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