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kkurt [141]
3 years ago
5

Sin(a-b) = sinacosb+cosasinb for all values of a and b. True or false?

Mathematics
2 answers:
vova2212 [387]3 years ago
8 0
It is a completely false statement that <span>Sin(a-b) = sinacosb+cosasinb for all values of a and b. The correct option among all the options that are given in the question is the second option. i hope that this is the answer that you were looking for and the answer has actually come to your desired help.</span>
guapka [62]3 years ago
7 0
False
The actual identity is:
Sin(a - b) = Sin(A)Cos(B) - Sin(B)Cos(A)
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Quadratic equations are second-order equations. Ginney made a mistake when she identified b = 31. It should be b = –31.

<h3>What is a Quadratic Equation?</h3>

A quadratic equation is an equation that can be written in the form of

ax²+bx+c.

Where a is the leading coefficient, and

c is the constant.

In order to find the mistake that Ginny made, we will first find the factors ourselves, therefore, we will factorize the trinomial,

6x^2 - 31x -30 = 0

As we can see that the value of constants are,

a = 6

b = -31

c = 30

6x^2 - 31x -30 = 0\\\\6x^2 - 36x+5x -30 = 0\\\\6x(x-6)+5(x-6)=0\\\\(6x+5)(x-6)=0

Therefore, the factors of the trinomial 6x^2 - 31x -30 = 0 are (6x+5) and (x-6).

If we compare our process with Ginny's process, we will find that she has taken the value of the constant b as 31 which is wrong.

Hence, Ginney made a mistake when she identified b = 31. It should be b = –31.

Learn more about Quadratic Equation:

brainly.com/question/17177510

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