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charle [14.2K]
3 years ago
8

Cos(2x)=1/4 first quadrant, find sinx

Mathematics
1 answer:
Shtirlitz [24]3 years ago
4 0
Solve is in the picture

Good luck :)

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3x²-9x+1+0 Find the discriminant
Sindrei [870]

Answer:

\boxed{D = 69}

Step-by-step explanation:

The given quadratic equation is:

3x^2-9x+1 = 0

Comparing it with the standard form of Quadratic Equation ax^2+bx+c = 0 , we get:

a = 3, b = -9 and c = 1

Discriminant = b² - 4ac

D = (-9)²-4(3)(1)

D = 81 - 12

D = 69

3 0
3 years ago
PLEASE HELP <br> it hasn’t been a good day and i don’t feel like doing this
Diano4ka-milaya [45]

Answer:

Both are quadrant

Step-by-step explanation:

4 0
3 years ago
Find the product.<br>3(m+ n)<br>3m+n<br>3m + 3n<br>3mn​
Fynjy0 [20]
You have to expand the expression so that it becomes
3m+3n
7 0
3 years ago
I need help will mark as BRAINIEST
Sladkaya [172]

Answer:

The answer is Linear absolute value .

I hope I helped ;-)

Step-by-step explanation:

4 0
3 years ago
A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha
Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

  • The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
  • the radius r=5cm
  • Punch is being poured into the bowl
  • The height at which the punch is increasing in the bowl is \frac{dh}{dt} = 1.5
  • the exposed area is a circle, (since the bowl is a hemisphere)
  • the radius of this circle can be written as 'a'
  • what is being asked is the rate of change of the exposed area when the height h = 2 cm
  • the rate of change of exposed area can be written as \frac{dA}{dt}.
  • since the exposed area is changing with respect to the height of punch. We can use the chain rule: \frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}
  • and since A = \pi a^2 the chain rule above can simplified to \frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt} -- we can call this Eq(1)

Solution:

the area of the exposed circle is

A =\pi a^2

the rate of change of this area can be, (using chain rule)

\frac{dA}{dt} = 2 \pi a \frac{da}{dt} we can call this Eq(2)

what we are really concerned about is how a changes as the punch is being poured into the bowl i.e \frac{da}{dh}

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

r = \frac{a^2 + h^2}{2h}

and rearrage the formula so that a is the subject:

a^2 = 2rh - h^2

now we can derivate a with respect to h to get \frac{da}{dh}

2a \frac{da}{dh} = 2r - 2h

simplify

\frac{da}{dh} = \frac{r-h}{a}

we can put this in Eq(1) in place of \frac{da}{dh}

\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}

and since we know \frac{dh}{dt} = 1.5

\frac{da}{dt} = \frac{(r-h)(1.5)}{a}

and now we use substitute this \frac{da}{dt}. in Eq(2)

\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}

simplify,

\frac{dA}{dt} = 3 \pi (r-h)

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

r = 5cm

h = 2cm from the question

\frac{dA}{dt} = 3 \pi (5-2)

\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s

5 0
3 years ago
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