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shusha [124]
1 year ago
12

Please help ! ASAP step by step explanation please ​

Mathematics
1 answer:
fredd [130]1 year ago
3 0

Answer:

0

Step-by-step explanation:

→ First find inverse cosine 1/2

60°

→ Now multiply this answer by 3 because then if you substitute it in you get 0.5

∝ = 180°

→ Now find sine of 180°

0

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Which two values of X are roots of the polynomial below 4x^2-6x+1
PtichkaEL [24]

4x^2-6x+1=0\\\\a=4,\ b=-6,\ c=1\\\\\Delta=b^2-4ac\\\\\Delta=(-6)^2-4(4)(1)=36-16=20\\\\\sqrt\Delta=\sqrt{20}=\sqrt{4\cdot5}=\sqrt4\cdot\sqrt5=2\sqrt5\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a},\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-(-6)-2\sqrt5}{2\cdot4}=\dfrac{3-\sqrt5}{4}\\\\x_2=\dfrac{-(-6)+2\sqrt5}{2\cdot4}=\dfrac{3+\sqrt5}{4}

5 0
3 years ago
A body moves s metres in a time t seconds so that s = t3 – 3t2 + 8. Find:
Lady bird [3.3K]

Using derivatives, it is found that:

i) v(t) = 3t^2 - 6t

ii) 9 m/s.

iii) a(t) = 6t - 6

iv) 6 m/s².

v) 1 second.

<h3>What is the role of derivatives in the relation between acceleration, velocity and position?</h3>
  • The velocity is the derivative of the position.
  • The acceleration is the derivative of the velocity.

In this problem, the position is:

s(t) = t^3 - 3t^2 + 8

item i:

Velocity is the <u>derivative of the position</u>, hence:

v(t) = 3t^2 - 6t

Item ii:

v(3) = 3(3)^2 - 6(3) = 27 - 18 = 9

The speed is of 9 m/s.

Item iii:

Derivative of the velocity, hence:

a(t) = 6t - 6

Item iv:

a(2) = 6(2) - 6 = 6

The acceleration is of 6 m/s².

Item v:

t for which a(t) = 0, hence:

6t - 6 = 0

6t = 6

t = \frac{6}{6}

t = 1

Hence 1 second.

You can learn more about derivatives at brainly.com/question/14800626

7 0
2 years ago
1 ) Considere a expressão algébrica:
balandron [24]
A) i think is correct andewmer
7 0
2 years ago
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