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Anastasy [175]
3 years ago
10

How many solutions does the system have? Explain. −3x + 6y = 10 −3x + 6y = −4

Mathematics
1 answer:
Temka [501]3 years ago
8 0

Answer:

none

Step-by-step explanation:

-3x + 6y cannot equal both 10 and -4 because 10 ≠ -4

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Answer:

1-c

2-b

3-a

Step-by-step explanation:

1. 7y=14

y=2

-7y=-14

y=2

so the equation has one solution

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so the equation has infinite solution

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3 years ago
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Luke grew some carrots in his garden. Their weights, in grams, are listed below.
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The bottom right histogram is the correct one.

Step-by-step explanation:

Carrots  weight 40-49 kg = 2

 ... of weight 50-59 = 6

 ...  of weight 60-69 = 4

 ..   of weight 70-79 = 1

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2 years ago
Some banks now use continuous compounding of an amount invested. In this case, the equation that models the value of an initial
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In this case, the equation that models the value of an initial investment of P dollars in t years at an annual interest rate of r is given by A = Pert.

Step-by-step explanation:

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2 years ago
2m = 1 + m what do I need to do to solve this problem
Anvisha [2.4K]

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3 years ago
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Find the exact value of each trigonometric function for the given angle θ.
Kay [80]

Answer:

\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since \theta=240, it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}

\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}

\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}

\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}

\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2

\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}

Therefore, \sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.

8 0
3 years ago
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