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Sati [7]
3 years ago
9

Help me with this question pleaseeee!!!

Mathematics
1 answer:
Butoxors [25]3 years ago
5 0
We have that
the measure of angle ∅=7*pi/6
the measure of its reference angle is?
remember that pi=180°
then 
7*pi/6=7*180/6=210°------------> III Quadrant
Part A) its reference angle is 210°-180°=30°
Part B) sin(30°)=-0.5--------> negative because  belongs III quadrant

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Simplify the following expression completely, where x
oksano4ka [1.4K]

Step-by-step explanation:

x \sqrt{5x {y}^{4} }  +  \sqrt{405 {x}^{3} y {}^{4} }  -  \sqrt{80 {x}^{3} {y}^{4}  }

x \sqrt{5xy {}^{4} }  + 9 x\sqrt{5xy {}^{4} }  - 4x \sqrt{5xy {}^{4} }

6x \sqrt{5xy {}^{4} }

8 0
3 years ago
Read 2 more answers
noel left home at twenty minutes to six and arrived at work twenty minutes past seven how long was his journel to work
Sati [7]

Answer:1 hour

Step-by-step explanation:

4 0
3 years ago
Use a system of equations to find the parabola of the form yequals=axsquared2plus+bxplus+c that goes through the three given poi
Andrei [34K]
Given that a parabola of the form y=ax^2+bx+c passing through the points (2, -11), (-2, -23) and (4, -53)

Thus, substituting the points we have:

-11=(2)^2a+2b+c \\  \\ \Rightarrow-11=4a+2b+c\ .\ .\ .\ (1) \\  \\ 
-23=(-2)^2a+(-2)b+c \\  \\ \Rightarrow -23=4a-2b+c\ .\ .\ .\ (2) \\  \\ 
-53=(4)^2a+4b+c \\  \\ \Rightarrow-53=16a+4b+c\ .\ .\ .\ (3)

We solve equations (1), (2) and (3) simulataneously. (There are many mays it can be solved but I will use row reduction method here).

We form the augumented matrix for equations (1), (2) and (3) and perform elementary row operations as follows:

\left[\begin{array}{ccc}4&2&1\\4&-2&1\\16&4&1\end{array}\right|  \left.\begin{array}{c}-11\\-23\\-53\end{array}\right]  \ \ \ \ \  \frac{1}{4} R_1\rightarrow R_1 \\  \\   \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\4&-2&1\\16&4&1\end{array}\right|  \left.\begin{array}{c}- \frac{11}{4} \\-23\\-53\end{array}\right]  \ \ \ \ \  {{-4R_1+R_2\rightarrow R_2} \atop {-16R_1+R_3\rightarrow R_3}}

\left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&-4&0\\0&-4&-3\end{array}\right|  \left.\begin{array}{c}- \frac{11}{4} \\-12\\-9\end{array}\right]  \ \ \ \ \  - \frac{1}{4} R_2  \\  \\  \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&1&0\\0&-4&-3\end{array}\right|  \left.\begin{array}{c}- \frac{11}{4} \\3\\-9\end{array}\right]  \ \ \ \ \   {{ -\frac{1}{2} R_2+R_1\rightarrow R_1} \atop {4R_2+R_3\rightarrow R_3}}

\left[\begin{array}{ccc}1&0& \frac{1}{4} \\0&1&0\\0&0&-3\end{array}\right|  \left.\begin{array}{c}- \frac{17}{4} \\3\\3\end{array}\right]  \ \ \ \ \  - \frac{1}{3} R_3 \\  \\  \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right|  \left.\begin{array}{c}-4\\3\\-1\end{array}\right]  \ \ \ \ \  - \frac{1}{4} R_3+R_1\rightarrow R_1

Thus, a = -4, b = 3, c = -1

Therefore, the required polynomial is y=-4x^2+3x-1
4 0
4 years ago
In the 2002 - 2003 NBA regular season, the Sacramento Kings
garri49 [273]

Answer: Number of games lost = 23 and Number of games won = 59.

Step-by-step explanation:

Let x = Number of games lost, y = Number of games won.

y = 13 + 2x    (i)

x+ y = 82    (ii)

Substitute value of y from (i) in (ii) , we get

x+ 13 + 2x  = 82

⇒ 3x= 82-13

⇒ 3x = 69

⇒  x= 23      [Divide both sides by 3]

Put value of x in (i), y= 13+2(23) = 13+46=59

Hence, Number of games lost = 23 and Number of games won = 59.

6 0
3 years ago
The present age of Joyal’s mother is three times the present age of joyal. After 5 years their ages will add up to 66. Find thei
bogdanovich [222]

。☆✼★ ━━━━━━━━━━━━━━  ☾  

Let's form some equations:

I'll refer to Joyal's age as 'x'

Therefore, his mother's age is '3x'

Here is the equation we create:

x + 5 + 3x + 5 = 66

Simplify this:

4x + 10 = 66

- 10 from both sides:

4x = 56

/4 on both sides

x = 14

3(14) = 42

So, currently Joyal is 14 and his mother is 42

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

7 0
3 years ago
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