Classify the system of equations 1/3x+y+5=0 1/2x+y-3=0

Mathematics

1 answer:

Pavel [41]3 years ago

6 0

Answer: Consistent and Independent.

Step-by-step explanation:

1. To solve this problem and classify the system of equations shown above, you can graph it has you can see in the graph attached.

2. As you can observe in the graph the lines intersect each other, this means that the system of equation has one solution.

3. By definition, when a system of equations has one solution, it is classified as Consistent and Independent.

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Plz use the graph help asap show work

Diano4ka-milaya [45]

Question 9.)
5-2(5)-1
10-1
Y=9
(X,Y) = 5,9
2- 2(2)-1
4-1
Y=3
(X,Y)= 2,3
0- 2(0)-1
2-1
Y=1
(X,Y)= 0,1
10)
4- 3(4)-7
12-7
Y=5
(X,Y)=4,3
1- 3(1)-7
3-7
Y=-4
(X,Y)= 1,-4
0- 3(0)-7
0-7
Y=-7
(X,Y)= 0,-7
(I believe you are supposed to plot the (X,Y) points on the graph)

7 0

3 years ago

The rate of change in sales for Garmin from 2008 through 2013 can be modeled by

Zina [86]

Answer:

a) The model for the sales of Garmin is represented by $S(t) = -\frac{81}{2500}\cdot t^{3} + \frac{267}{250}\cdot t^{2} - 11.9\cdot t + 47.112$ .

b) The average sales of Garmin from 2008 through 2013 were $ 2.5 billion.

Step-by-step explanation:

a) The model for the sales of Garmin is obtained by integration:

$S(t) = -0.0972\int {t^{2}} \, dt + 2.136\int {t}\,dt -11.9 \int\,dt$

$S(t) = -\frac{81}{2500}\cdot t^{3} + \frac{267}{250}\cdot t^{2} - 11.9\cdot t + C$ (1)

Where $C$ is the integration constant.

If we know that $t = 9$ and $S(t) = 2.9$ , then the model for the sales of Garmin is:

$-\frac{81}{2500} \cdot 9^{3} + \frac{267}{250}\cdot 9^{2}-11.9\cdot (9) + C = 2.9$

$C = 47.112$

The model for the sales of Garmin is represented by $S(t) = -\frac{81}{2500}\cdot t^{3} + \frac{267}{250}\cdot t^{2} - 11.9\cdot t + 47.112$ .

b) The average sales of the Garmin from 2008 through 2013 ( $\bar{S}$ ) is determined by the integral form of the definition of average, this is:

$\bar{S} = \frac{1}{13 - 8} \cdot \int\limits^{13}_{8} {S(t)} \, dt$ (2)

$\bar S = \frac{1}{5}\cdot \int\limits^{13}_{8} {\left[-\frac{81}{2500}\cdot t^{3} + \frac{267}{250}\cdot t^{2}-11.9\cdot t + 47.112 \right]} \, dt$

$\bar S = \frac{1}{5}\cdot \left[-\frac{81}{10000}\cdot (13^{4}-8^{4}) +\frac{89}{250}\cdot (13^{3}-8^{3}) -\frac{119}{20}\cdot (13^{2}-8^{2}) +47.112\cdot (13-8) \right]$ $\bar{S} = \frac{1}{5}\cdot (-198.167+599.86-624.75+235.56)$