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Orlov [11]
3 years ago
15

Which of the following statements are true of the

Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

B, D, and E.

Step-by-step explanation:

A) The system has infinitely many solutions. This is wrong because according to the graph, there is only one solution- where the lines intersect. This would only be true if the lines never intersected.

B) A solution to the system is (-1, -2). This is true because this is the only point where the lines intersect.

C) A solution to the system is (0, -1). Since these aren't parabolas and the one above is true, we can say this is false. Also, the lines don't intersect at (0, -1).

D) One of the equations is y=x-1. This is true because the y-intercept for the red line is -1 and the slope of the equation is 1. You can also find this out by directly solving for the equation.

E) One of the equations is 3x+y=-5. If you put this into slope-intercept form, you will find out that the equation is y=-3x-5. This is true because the y-intercept of this is -5 and the slope of this is -3.

andreev551 [17]3 years ago
8 0

Answer:

B,D and E

Step-by-step explanation:

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2 years ago
Find the unit vector in the direction of u = (-3,2).
DENIUS [597]
\bf \textit{unit vector for }(a,b)\implies \left( \cfrac{a}{\sqrt{a^2+b^2}}~~,~~\cfrac{b}{\sqrt{a^2+b^2}} \right)\\\\
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(-3,2)\qquad \stackrel{unit~vector}{\implies }\qquad \left( \cfrac{-3}{\sqrt{(-3)^2+2^2}}~~,~~\cfrac{2}{\sqrt{(-3)^2+2^2}} \right)
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\left( -\cfrac{3}{\sqrt{13}}~~,~~ \cfrac{2}{\sqrt{13}}\right)\\\\
-------------------------------

\bf \textit{and now let's \underline{rationalize} the denominator for each}
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-\cfrac{3}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies -\cfrac{3\sqrt{13}}{13} \qquad \qquad \qquad \qquad  \cfrac{2}{\sqrt{13}}\cdot \cfrac{\sqrt{13}}{\sqrt{13}}\implies \cfrac{2\sqrt{13}}{13}
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\textit{and written in \underline{ai+bj form}}\qquad -\cfrac{3\sqrt{13}}{13}i~~~~+~~~~\cfrac{2\sqrt{13}}{13}j
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3 years ago
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