What is the answer to the q's

What is the frequency of the sinusoidal graph?

sleet_krkn [62]

Answer:

$\frac{3}{2\pi }$

Step-by-step explanation:

3 0

3 years ago

TRIANGLE CONGRUENCE! PICTURE ADDED. How is QTP congruent to SPT?

shtirl [24]

Answer:

The triangles are congruent because:

QP = ST
angle P = angle T
the side PT is in both triangles

We know when 2 sides of two triangles and the angle between them is congruent, the triangles are congruent.

5 0

3 years ago

Solve the equation for h? X=r-h/y

Troyanec [42]

Answer:

h = yX -yr

Step-by-step explanation:

X=r-h/y

X - r = h/y

y*(X-r) = h

h = yX -yr

4 0

3 years ago

A colony of flowers is pollinated by hummingbirds and sunbirds. Let H represents the number of hummingbirds and S represent the

zloy xaker [14]

Complete Question:

A colony of flowers is pollinated by hummingbirds and sunbirds. Let H represents the number of hummingbirds and S represent the number of sunbirds that must pollinate the colony so it can survive until next year.

6H+ 4S> 74

This year, 8 hummingbirds pollinated the colony. What is the least number of sun birds that must pollinate the colony to ensure that it will survive until next year?

Answer:

The least number of sun birds is 7

Solution:

Given inequality is:

6H + 4S > 74

Where,

H is the number of hummingbirds

S is the number of sunbirds

This year, 8 hummingbirds pollinated the colony

Substitute H = 8 in given inequality

6(8) + 4S > 74

48 + 4S > 74

Subtract 48 from both sides

48 + 4S - 48 > 74 - 48

4S > 26

Divide both sides by 4

S > 6.5

Hence the least number of sun birds is 7

7 0

3 years ago

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

3 years ago