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4 years ago

Write the expression as the sine, cosine, or tangent of an angle. cos 107° cos 42° + sin 107° sin 42°

2 answers:

8 0

Let a =107° and b=42°;

So cos 107° cos 42° + sin 107° sin 42° = cosa.cos42 + sina.sinb = cos(a-b)

and cos 107° cos 42° + sin 107° sin 42° = cos(107-42) = cos(65°)

miskamm [114]4 years ago

8 0

Answer: It will be

$\cos65\textdegree\ and\ \sin 25\textdegree$

Step-by-step explanation:

Since we have given that

cos 107° cos 42° + sin 107° sin 42°

As we know the formula :

$\cos A\cos B+\sin A\sin B=cos(A-B)$

So, applying the above formula , we get

$A=107\textdegree\\B=42\textdegree$

Now, we have to write it in cosine terms,

So, we get,

$\cos(107\textdegree-42\textdegree)=\cos65\textdegree$

So, in terms of cosine , it will be

$\cos65\textdegree$

In terms of sine , it will be

$\cos 65\textdegree=\sin(90\textdegree-65\textdegree)=\sin 25\textdegree$

Hence, it will be

$\cos65\textdegree\ and\ \sin 25\textdegree$

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An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is

vladimir1956 [14]

Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

b) $P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

c) $P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

Part b

We want this probability:

$P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

We find the individual probabilities like this:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

Part c

For this case we want this probability:

$P(X \geq 5)$

And we can use the complement rule like this:

$P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

3 0

3 years ago

How to do Simplification of algebraic fractions

Marysya12 [62]

If we want to simplify Algebraic Fractions:
We have to take a look at the numerator and the denominator
Why? Because to simplify we need to have a common multiple.
When we find the common multiple, we now know by which factor we have to break down the fraction in the numerator and the denominator
I have some examples attached below for help.