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Gala2k [10]
3 years ago
11

AI Prove that:1-cOS AV1+cos Acosec A-cot A​

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
7 0

Answer:

(1-cosA)/(1+cosA)

=(1-cosA)/(1+cosA) ×(1-cosA)/(1-cosA)

=(1-cosA)^2 /(1-cos^2A)

=(1-cosA)^2 / (sin^2A)

=(1/sinA - cosA/sinA )^2

=cosecA - cotA)^2

Hope it helps

Have a great Day ; )

Dovator [93]3 years ago
6 0

Answer:

at least can I get Ur g m a i l plz

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Alekssandra [29.7K]

Answer:

-1/2

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Write your answer in simplest radical form​
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z = √3

Step-by-step explanation:

sin (30°) = z / 2√3

z = sin (30°) 2√3

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To solve -8p = 48, which of the following could you do to both sides of the equation? A. add -8 B. subtract -8 C. multiply by -8
masya89 [10]

Answer:

D. divide by -8

Step-by-step explanation:

8 0
3 years ago
getting home from trick or treat celia and emma counted their candies. half of celias candies is equal to 2/3 of emmas candies.
blagie [28]

Candies with celias and emmas is 60 and 45 respectively.

<u>Solution:</u>

Given, Getting home from trick or treat celia and emma counted their candies.  Half of celias candies is equal to 2/3 of emmas candies.  

They had a total of 105 candies altogether.  

We have to find how many candies did each of them have.

Let the number of candies with celias be n, then number of candies with emma will be 105 – n.

Now according to given condition.

\begin{array}{l}{\frac{1}{2} \times \text { celias candies count }=\frac{2}{3} \times \text { emmas candies count }} \\\\ {\rightarrow \frac{1}{2} \times n=\frac{2}{3} \times(105-n)} \\\\ {\rightarrow 3 \times n=2 \times 2 \times(105-n)} \\\\ {\rightarrow 3 \times n=2 \times 2 \times(105-n)} \\\\ {\rightarrow 3 n=4(105-n)} \\\\ {\quad \rightarrow 3 n=420-4 n} \\\\ {\rightarrow 3 n=420-4 n} \\\\ {\rightarrow 3 n+4 n=420} \\\\ {\rightarrow 7 n=7 \times 60} \\\\ {\rightarrow n=60}\end{array}

Hence, candies with celias and emmas is 60 and 45 respectively.

4 0
3 years ago
Please show working out thanks
IgorLugansk [536]

Answer:

10m x 15m

Step-by-step explanation:

You are given some information.

1. The area of the garden: A₁ = 150m²

2. The area of the path: A₂ = 186m²

3. The width of the path: 3m

If the garden has width w and length l, the area of the garden is:

(1) A₁ = l * w

The area of the path is given by:

(2) A₂ = 3l + 3l + 3w + 3w + 4*3*3 = 6l + 6w + 36

Multiplying (2) with l gives:

(3) A₂l = 6l² + 6lw + 36l

Replacing l*w in (3) with A₁ from (1):

(4) A₂l = 6l² + 6A₁ + 36l

Combining:

(5) 6l² + (36 - A₂)l +6A₁ = 0

Simplifying:

(6) l² - 25l + 150 = 0

This equation can be factored:

(7) (l - 10)*(l - 15) = 0

Solving for l we get 2 solutions:

l₁ = 10, l₂ = 15

Using (1) to find w:

w₁ = 15, w₂ = 10

The two solutions are equivalent. The garden has dimensions 10m and 15m.

3 0
3 years ago
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