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Sophie [7]
3 years ago
9

Find the reference angle given: t = -216º.

Mathematics
1 answer:
Mandarinka [93]3 years ago
5 0

I not sure but I think it's 180°

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X^2 + 5x = quadratic equation <br>=​
timama [110]

Hi there!

The question gives us the quadratic equation , and it tells us to solve it using the quadratic formula, which goes as . However, we must first find the values of a, b, and c. The official quadratic equation goes as , which matches the format of the given quadratic equation. Hence, the value of a would be 1, the value of b would be 5, and the value of c would be 3. Now, just plug it back into the quadratic equation and simplify to get the zeros of the equation. 

x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}  

x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(3)} }{2(1)}  

x = \frac{-5 \pm \sqrt{25 - 12} }{2}  

x = \frac{-5 \pm \sqrt{13} }{2}  

x = \frac{-5 \pm 3.61 }{2}  

x = \frac{-5 + 3.61 }{2}, x = \frac{-5 - 3.61 }{2}

x=-0.695 \ \textgreater \ \ \textgreater \  -0.7, x= -4.305 \ \textgreater \ \ \textgreater \ x=-4.31

Therefore, the solutions to the quadratic equation  are x = -0.7 and x = -4.31. Hope this helped and have a phenomenal day!

Your answer is 4.31

4 0
3 years ago
At 6:00 a.m., the outside temperature was –14.4⁰. By noon, the outside temperature had dropped 8.1⁰. What was the outside temper
Advocard [28]
-14.4 - 8.1 = - 22.5

The outside temperature at noon is - 22.5 degrees
5 0
2 years ago
Read 2 more answers
HELLPPP GRADESSS AHAHHAH [brainly
Umnica [9.8K]

Answer: 17x+9

Step-by-step explanation:

6 0
2 years ago
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A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a
AnnZ [28]
  <span>Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed: 
a = omega^2*r 

omega and r in terms of given data: 
omega = 2*Pi/T 
r = d/2 

Thus: 
a = 2*Pi^2*d/T^2 

What forces cause this acceleration for the passenger, at either top or bottom? 

At top (acceleration is downward): 
Weight (m*g): downward 
Normal force (Ntop): upward 

Thus Newton's 2nd law reads: 
m*g - Ntop = m*a 

At top (acceleration is upward): 
Weight (m*g): downward 
Normal force (Nbottom): upward 

Thus Newton's 2nd law reads: 
Nbottom - m*g = m*a 

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame: 
Ntop = m*(g - a) 
Nbottom = m*(g + a) 


Substitute a: 
Ntop = m*(g - 2*Pi^2*d/T^2) 
Nbottom = m*(g + 2*Pi^2*d/T^2) 

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground): 
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g) 
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g) 

Simplify: 
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2) 
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2) 

Data: 
d:=22 m; T:=12.5 sec; g:=9.8 N/kg; 

Results: 
Ntop/(m*g) = 71.64%...she feels "light" 
Nbottom/(m*g) = 128.4%...she feels "heavy"</span>
7 0
3 years ago
Read 2 more answers
Please help me with this!!
nydimaria [60]

Answer:

Step-by-step explanation:

these triangles are not similar,  you can tell, b/c the one given angle and that they are both right triangles, so you also know that the other angle , that is not given is not the same, in the two.   hence the sides are not the same lengths or ratios of each other.  they are NA

6 0
2 years ago
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