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Elan Coil [88]
3 years ago
13

Verify the identity: cos(a+B)/cos a sin B = cot B -tan a

Mathematics
1 answer:
alexgriva [62]3 years ago
5 0
Cos( A + B ) = cosAcosB - sinAsinB ;
cos( A + B ) / ( cosAsinB ) = ( cosAcosB - sinAsinB ) / ( cosAsinB ) = ( cosAcosB ) / ( cosAsinB ) - ( sinAsinB ) / ( cosAsinB ) = cosB / sinB - sinA / cosA = cotB - tanA ;
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Investment of $3000 at 16% simple interest for two years.
Yanka [14]
I=PRT
r=rat
P=present amount
I=interst
t=time in yeare

2 years
16%
3000
I=2*3000*0.16
I=$960

3 0
3 years ago
Read 2 more answers
Mr. Newman works as an architect. He created a solid
Andrews [41]

Answer: The answer is 34,000 cm

Step-by-step explanation: I just did that

6 0
3 years ago
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What is the answer to -6n^2 - 5n - 7 = -8
soldi70 [24.7K]

Answer:

n = \frac{1}{3} , n = \frac{1}{2}

Step-by-step explanation:

6n² - 5n - 7 = - 8 ( add 8 to both sides )

6n² - 5n + 1 = 0 ← in standard form

Consider the product of the factors of the coefficient of the n² term and the constant term which sum to give the coefficient of the n- term

product = 6 × 1 = 6 and sum = - 5

The factors are - 3 and - 2

Use these factors to split the n- term

6n² - 3n - 2n + 1 = 0 ( factor the first/second and third/fourth terms )

3n(2n - 1) - 1(2n - 1) = 0 ← factor out (2n - 1) from each term

(2n - 1)(3n - 1) = 0 ← in factored form

Equate each factor to zero and solve for n

3n - 1 = 0 ⇒ 3n = 1 ⇒ n = \frac{1}{3}

2n - 1 = 0 ⇒ 2n = 1 ⇒ n = \frac{1}{2}

7 0
3 years ago
PLEASE HELP ME OUT ASAP!!! GEOMETRY! :V
Dima020 [189]
For the fish bowl the first answer is correct.
(1/2)(4/3)(22/7)(12^3)

For the second problem
(1/3)(3.14)(6^2)(10) - (4/3)(3.14)(1.5^3)
The second answer.
5 0
3 years ago
What’s the vertex form of F(x)=x^2+2x-3
kupik [55]

Answer:

\large\boxed{f(x)=(x+1)^2-4}

Step-by-step explanation:

\text{The vertex form of a quadratic equation}\ f(x)=ax^2+bx+c:\\\\f(x)=a(x-h)^2+k\\\\(h,\ k)-\text{vertex}\\=====================================

\bold{METHOD\ 1:}\\\\\text{convert to the perfect square}\ (a+b)^2=a^2+2ab+b^2\qquad(*)\\\\f(x)=x^2+2x-3=\underbrace{x^2+2(x)(1)+1^2}_{(*)}-1^2-3\\\\f(x)=(x+1)^2-4\\==============================

\bold{METHOD\ 2:}\\\\\text{Use the formulas:}\ h=\dfrac{-b}{2a},\ k=f(k)\\\\f(x)=x^2+2x-3\to a=1,\ b=2,\ c=-3\\\\h=\dfrac{-2}{2(1)}=\dfrac{-2}{2}=-1\\\\k=f(-1)=(-1)^2+2(-1)-3=1-2-3=-4\\\\f(x)=(x-(-1))^2-4=(x+1)^2-4

4 0
3 years ago
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