Prove that 1-tan32°tan13°-tan32°-tan13°=0

Mathematics

1 answer:

Step2247 [10]2 years ago

3 0

Answer:

Step-by-step explanation:

45=32+13

tan 45=tan(32+13)

$1=\frac{tan~ 32+tan~ 13}{1-tan ~32 ~tan~ 13} \\cross~multiply \\1-tan 32 ~tan~13=tan ~32+tan~13\\$

or 1-tan 32 tan 13-tan 32-tan 13=0

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Let f(x)=2.912345x^2+3.131579x-0.099999

Rudiy27

A.) Integers are positive and negative counting numbers. So, in order to find the integer coefficients, round off the coefficients in the equation to the nearest whole number. The function for g(x) is:

g(x) = 3x²+3x

B.) Substitute x=4 to the two functions.

f(x) = 2.912345x²+3.131579x-0.099999
f(4) = 2.912345(4)²+3.131579(4)-0.099999
f(4) = 59.023837

g(x) = 3x²+3x
g(4) = 3(4)²+3(4)
g(4) = 60

C.) The percentage error is equal to:

Percentage error = |g(4) - f(4)|/f(4) * 100
Percentage error = |60 - 59.023837|/59.023837 * 100
Percentage error = 1.65%

D.) If x is a large number, for example x=10 or x=20, then g(x) would be an overestimate. This is because the value of x is raised to the power of 2. So, as the x increases, the corresponding function would increase exponentially. Even at x=4, g(x) is already an overestimate. What more for larger values of x? That means that the gap from the true answer f(x) would increase.

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3 years ago

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IRISSAK [1]

The last 2 options are correct

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3 years ago

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For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure.

Finger [1]

Answer:

Lower limit: 113.28

Upper limit: 126.72

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 120, \sigma = 8$