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PSYCHO15rus [73]
3 years ago
8

find three consecutive even integers so that the twice the sum of the second and third is twelve less than six times the first

Mathematics
1 answer:
zysi [14]3 years ago
4 0

n, n + 2, n + 4 - three consecutive even integers

the twice the sum of the second and third: 2[(n + 2) + (n + 4)]

twelve less than six times the first: 6n - 12

The equation:

2[(n + 2) + (n + 4)] = 6n - 12

2(n + 2 + n + 4) = 6n - 12

2(2n + 6) = 6n - 12       <em>use distributive property</em>

(2)(2n) + (2)(6) = 6n - 12

4n + 12 = 6n - 12       <em>subtract 12 from both sides</em>

4n = 6n - 24       <em>subtract 6n from both sides</em>

-2n = -24        <em>divide both sides by (-2)</em>

n = 12

n + 2 = 12 + 2 = 14

n + 4 = 12 + 4 = 16

<h3>Answer: 12, 14, 16</h3>
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What is the solution to the system of equations y=-5x+3
Nimfa-mama [501]

Hola User______

Here is Your Answer...!!!!

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thus the given equation is a linear equation in two variables ...

thus ploting a table we get as ...

y = -5x + 3

there fore

1) at y=0 , x = 3/5 .... 0 = -5x +3

= -5x = -3 ..thus X = 3/5

2) at x=0 , y= 3 ...

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4 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

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1 year ago
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Answer:

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3 years ago
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