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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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Find the Greatest Common Factorthe largest number that divides evenly into <span>4x<span>4x</span></span> and <span>-6<span>−6</span></span>?
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the highest degree of <span>xx</span> that divides evenly into <span>4x<span>4x</span></span> and <span>-6<span>−6</span></span>?
It is 1, since <span>xx</span> is not in every term.Multiplying the results above,
The GCF is <span>22</span>.

<span><span>2(<span><span>​2</span>​<span>​<span>4x</span></span><span>​​</span></span>−<span><span>​2</span>​<span>​6</span><span>​​</span></span>)

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3 0
2 years ago
QUESTION IN THE ATTACHMENT
eimsori [14]

Answer:

A. The sum of the first 10th term is 100.

B. The sum of the nth term is n²

Step-by-step explanation:

Data obtained from the question include:

Sum of 20th term (S20) = 400

Sum of 40th term (S40) = 1600

Sum of 10th term (S10) =..?

Sum of nth term (Sn) =..?

Recall:

Sn = n/2[2a + (n – 1)d]

Sn is the sum of the nth term.

n is the number of term.

a is the first term.

d is the common difference

We'll begin by calculating the first term and the common difference. This is illustrated below:

Sn = n/2 [2a + (n – 1)d]

S20 = 20/2 [2a + (20 – 1)d]

S20= 10 [2a + 19d]

S20 = 20a + 190d

But:

S20 = 400

400 = 20a + 190d .......(1)

S40 = 40/2 [2a + (40 – 1)d]

S40 = 20 [2a + 39d]

S40 = 40a + 780d

But

S40 = 1600

1600 = 40a + 780d....... (2)

400 = 20a + 190d .......(1)

1600 = 40a + 780d....... (2)

Solve by elimination method

Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:

40 x equation 1:

40 x (400 = 20a + 190d)

16000 = 800a + 7600. ........ (3)

20 x equation 2:

20 x (1600 = 40a + 780d)

32000 = 800a + 15600d......... (4)

Subtract equation 3 from equation 4

Equation 4 – Equation 3

32000 = 800a + 15600d

– 16000 = 800a + 7600d

16000 = 8000d

Divide both side by 8000

d = 16000/8000

d = 2

Substituting the value of d into equation 1

400 = 20a + 190d

d = 2

400 = 20a + (190 x 2)

400 = 20a + 380

Collect like terms

400 – 380 = 20a

20 = 20a

Divide both side by 20

a = 20/20

a = 1

Therefore,

First term (a) = 1.

Common difference (d) = 2.

A. Determination of the sum of the 10th term.

First term (a) = 1.

Common difference (d) = 2

Number of term (n) = 10

Sum of 10th term (S10) =..?

Sn = n/2 [2a + (n – 1)d]

S10 = 10/2 [2x1 + (10 – 1)2]

S10 = 5 [2 + 9x2]

S10 = 5 [2 + 18]

S10 = 5 x 20

S10 = 100

Therefore, the sum of the first 10th term is 100.

B. Determination of the sum of the nth term.

First term (a) = 1.

Common difference (d) = 2

Sum of nth term (Sn) =..?

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [2x1 + (n – 1)2]

Sn = n/2 [2 + 2n – 2]

Sn = n/2 [2 – 2 + 2n ]

Sn = n/2 [ 2n ]

Sn = n²

Therefore, the sum of the nth term is n²

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