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viktelen [127]
3 years ago
10

Jason drew a right triangle with complementary angles that measure (x + 3)° and (x + 9)°. What are the measures of the complemen

tary angles?
A. 44° and 46°
B. 89° and 91°
C. 42° and 48°
D. 51° and 39°
Mathematics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

C. 42° and 48°

Step-by-step explanation:

Complementary angles: Angles that add up to 90°.

So we can set up an equation..

(x + 3) + (x + 9) = 90

2x + 12 = 90

2x = 78

x = 39°

Now we can find the angles.

Angle 1:

(x + 3)

= 39 + 3

= 42°

Angle 2:

(x + 9)

= 39 + 9

= 48°

And we can check to see if they add to 90°,

42° + 48° = 90°

         

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The solution for l4  is  mathematically given as

L_{4}=0.5431

<h3>What is the solution for l4?</h3>

&\text { Given } f(x)=a \cos ^{2}(x) \quad\left[\frac{\pi}{8}, \frac{\pi}{2}\right] \text { \& } n=4\\\\&\Delta a=\frac{b-a}{n}=\frac{\pi / 2-\pi / 8}{9}=\frac{3 \pi}{32}\\\\&x_{0}=\pi / 8, x_{1}=\frac{\pi}{8}+\frac{3 \pi}{32}=\frac{7 \pi}{32}\\\\&x_{2}=\frac{5 \pi}{16}, \quad x_{3}=\frac{13 \pi}{32}, x_{4}=\pi / 2\\\\&f\left(x_{0}\right)=f(1 / 8)=0.8535\\

&f\left(x_{1}\right)=f\left(\frac{7 \pi}{32}\right)=0.5975\\\\\&f\left(x_{2}\right)=f\left(\frac{5 \pi}{16}\right)=0.3086\\\\\&f\left(a_{3}\right)=f\left(\frac{13 \pi}{32}\right)=0.0842\\\\\&L_{4}=\sum_{k_{0}^{=0}}^{3} f\left(x_{k}\right) \Delta x\\\\&=\Delta \\\\x\left[f\left(x_{0}\right)+f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)\right]\\\\\&=\frac{3 \pi}{32}[0.8535+0.5975+0.3056+0.0842]\\\\&L_{4}=0.5431

Read more about  approximation

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The complete Question is attached below

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Step-by-step explanation:

<h3>Question-1:</h3>

by order pair we obtain:

\displaystyle  \begin{cases}   \displaystyle 3p = 2p - 1 \dots \dots i\\2q - p =  1 \dots  \dots ii\end{cases}

cancel 2p from the i equation to get a certain value of p:

\displaystyle  \begin{cases}   \displaystyle p =  - 1 \\2q - p =  1 \end{cases}

now substitute the value of p to the second equation:

\displaystyle  \begin{cases}   \displaystyle p =  - 1 \\2q - ( - 1) =  1 \end{cases}

simplify parentheses:

\displaystyle  \begin{cases}   \displaystyle p =  - 1 \\2q  +  1=  1 \end{cases}

cancel 1 from both sides:

\displaystyle  \begin{cases}   \displaystyle p =  - 1 \\2q  =  0\end{cases}

divide both sides by 2:

\displaystyle  \begin{cases}   \displaystyle p =  - 1 \\q  =  0\end{cases}

<h3>question-2:</h3>

by order pair we obtain:

\displaystyle  \begin{cases}   \displaystyle 2x - y= 3 \dots \dots i\\3y= x + y \dots  \dots ii\end{cases}

cancel out y from the second equation:

\displaystyle  \begin{cases}   \displaystyle 2x - y= 3 \dots \dots i\\ x = 2y \dots  \dots ii\end{cases}

substitute the value of x to the first equation:

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simplify:

\displaystyle  \begin{cases}   \displaystyle 3y= 3 \\ x = 2y \end{cases}

divide both sides by 3:

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substitute the value of y to the second equation which yields:

\displaystyle  \begin{cases}   \displaystyle y= 1 \\ x = 2 \end{cases}

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by order pair we obtain;

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rearrange:

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hence,

\displaystyle  q =  \frac{1}{2} \\ p =   \frac{3}{4}

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