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VladimirAG [237]

3 years ago

12

Determine the minimum and maximum for y=400 sin(2t)+1200

1 answer:

natta225 [31]3 years ago

7 0

The maximum of sin(2t) is 1
and the minimum of it is -1

so the maximum of y = 400*1 + 1200 = 1600
and the minimum of y = 400*(-1) + 1200 = 800

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The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in <em>fourth</em> quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in <em>4th quadrant </em>so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago

Can someone pls help me i really don't know how to do this

Damm [24]

Answer:

option 2

Step-by-step explanation:

Since the sign is greater than or equal to it will be pointing to the right since it includes -9 it will be a closed circle

7 0

3 years ago

The ratio of the width to the length of a rectangle is 2:3, respectively. Answer each of the following. a By what percent would

Elina [12.6K]

Answer:

The percentage increase in Area of rectangle is 200%

Step-by-step explanation:

Given as :

The ratio of the width to the length of a rectangle is 2:3

Let The length of rectangle = 2 x

Let The width of rectangle = 3 x

∵ Area of rectangle = length × width

So, $A_1$ = 2 x × 3 x

Or, $A_1$ = 6 x²

<u>Again</u>

The increased length of rectangle = 2 x + 2 x = 4 x

The increase width of rectangle = 3 x + 50% of 3 x

I.e The increase width of rectangle = 3 x + 1.5 x = 4.5 x

∵ Increased Area of rectangle = increased length × increased width

Or, $A_2$ = 4 x × 4.5 x = 18 x²

So, The percentage increase in area = $\dfrac{A_2 - A_1}{A_1}$ × 100

Or, The percentage increase in area = $\frac{18 x^{2}-6x^{2}}{6x^{2}}$ × 100

Or , The percentage increase in area = $\dfrac{12}{6}$ × 100

∴ The percentage increase in area = 200

Hence, The percentage increase in Area of rectangle is 200% Answer

5 0

3 years ago

Solve for x. Round your answer to the nearest tenth.

inna [77]

Answer is 360,,,,,,,,,........

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3 years ago

What is 5.823 rounded to the nearest whole<br> what is 3.147 rounded to the nearest whole

stich3 [128]

Answer:

6 and 3

Step-by-step explanation:

5.8 to the nearest whole is 6

3.147 to the nearest whole is 3

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2 years ago