MATHEMATICS MIDDLE SCHOOL

Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:y < 4x − 8
y is greater than or equal to negative 5 over 2 times x plus 5

Part A: Describe the graph of the system, including shading and the types of lines graphed. Provide a description of the solution area. (6 points)

Part B: Is the point (5, −8) included in the solution area for the system? Justify your answer mathematically. (4 points)

Answers

Answer 1

Answer:

Answer:

:Part A) The graph in the attached figure (see the explanation)

Part B) The ordered pair is not included in the solution area for the system (see the explanation)

Step-by-step explanation:

Part A) we have

$y<4x-8$ ----> inequality A

The solution of the inequality A is the shaded area below the dashed line $y=4x-8$

The slope of the dashed line is positive

The y-intercept of the dashed line is (0,-8)

The x-intercept of the dashed line is (2,0)

$y\geq -(5)/(2)x+5$ ----> inequality B

The solution of the inequality B is the shaded area above the solid line $y=-(5)/(2)x+5$

The slope of the solid line is negative

The y-intercept of the solid line is (0,5)

The x-intercept of the solid line is (2,0)

The solution of the system of inequalities is the shaded area between the dashed line and the solid line

see the attached figure

Part B) is the point (5, -8) Included in the solution area for the system?

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Substitute the value of x=5, y=-8 in each inequality and then analyze the results

Inequality A

$-8<4(5)-8$

$-8<12$ ----> is true

the ordered pair satisfy inequality A

Inequality B

$-8\geq -(5)/(2)(5)+5$

$-8\geq -7.5$ ---> is not true

the ordered pair not satisfy the inequality B

therefore

The ordered pair is not included in the solution area for the system

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Answers

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b) at least 75 papers will be missing the name
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Answers

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The edge of one cube is 4 m shorter than the edge of a second cube. The volumes of the two cubes differ by 1216 m^3. Find the edge of the smaller cube.

Answers

Volume of cube, V = edge^3
Let edge of cube#1 = (x-4) m, therefore volume of cube#1, v1 = (x-4)^3 m
Let edge of cube#2 = x m, therefore volume of cube#2, v2 = x^3 m
Diff. in volume (in m) = 1216 = v2-v1 = [ x^3 - (x-4)^3 ]
= x^3 - [(x-4)(x-4)(x-4)]
= x^3 - [x^2 - 8x +16(x - 4)]
= x^3 - [ x^3 - 12x^2 + 48x - 64 ]
= 12x^2 - 48x + 64
= 4 (3x^2 - 12x + 16)
Therefore 4 (3^2 - 12x + 16) = 1216
3x^2 - 12x + 16 = 1216/4 = 304
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3 (x^2 - 4x - 96) = 0
(x^2 - 4x - 96) = 0
(x - 12) (x + 8) =0
(x-12) = 0
Therefore x = 12 m
Edge of cube#2 = x m = 12m
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$a- the\ edge\ of\ the\ smaller\ cube\ \ \ \wedge\ \ \ a>0\nV_a-the\ volume\ of\ the\ smaller\ cube\ \ \ \Rightarrow\ \ \ V_a=a^3\n \n(a+4)^3-a^3=1216\ \ \ \wedge\ \ \ (x+y)^3=x^3+3x^2\cdot y+3x\cdot y^2+y^3\n \na^3+3a^2\cdot 4+3a\cdot 4^2+4^3-a^3=1216\n \n12a^2+48a+64-1216=0\ \ \ \Rightarrow\ \ \ 12a^2+48a-1152=0\ /:12\n \na^2-4a-96=0\ \ \ \Rightarrow\ \ \ \Delta=(-4)^2-4\cdot 1\cdot(-96)=16+384=400\n \n √(\Delta) =20\n \na_1= (4-20)/(2\cdot1) =-8<0,\ \ \ a_2= (4+20)/(2\cdot1) =12>0$

$Ans.\ The\ edge\ of\ the\ smaller\ cube=12\ m$

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Answers

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Answers

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