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2 years ago

Please help pleaseeeeeee

Mathematics

2 answers:

Viktor [21]2 years ago

6 0

Answer:

it would be C

Step-by-step explanation:

marysya [2.9K]2 years ago

3 0

Answer:

Step-by-step explanation:

it's basically 2+7

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skelet666 [1.2K]

Hi! it's the third statement

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3 years ago

(08.02)Which of the following graphs best represents the solution to the pair of equations below?

AnnZ [28]

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Step-by-step explanation:

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2 years ago

A gardener has 520 feet of fencing to fence in a rectangular garden. One side of the garden is bordered by a river and so it doe

Bond [772]

The shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet

<h3>What dimensions would guarantee that the garden has the greatest possible area?</h3>

The given parameter is

Perimeter, P = 520 feet

Represent the shorter side with x and the longer side with y

One side of the garden is bordered by a river:

So the perimeter is:

P = 2x + y

Substitute P = 520

2x + y = 520

Make y the subject

y = 520 - 2x

The area is

A = xy

Substitute y = 520 - 2x in A = xy

A = x(520 - 2x)

Expand

A = 520x - 2x^2

Differentiate

A' = 520 - 4x

Set to 0

520 - 4x = 0

Rewrite as:

4x= 520

Divide by 4

x= 130

Substitute x= 130 in y = 520 - 2x

y = 520 - 2 *130

Evaluate

y = 260

The area is then calculated as:

A = xy

This gives

A = 130 * 260

Evaluate

A = 33800

Hence, the shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet

Read more about area at:

brainly.com/question/24487155

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1 year ago

Vanessa's scale drawing of a volleyball court, shown below, is 3 inches long and 1 1/2 inches wide. How many square meters make

natima [27]

I took a test on this yesterday and all i know is the answer is either 81 117 or 162

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3 years ago

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

2 years ago