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dedylja [7]
3 years ago
11

Trig question please save mee simplify each expression to one trig function or number

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
5 0

Answer:

sec x

Step-by-step explanation:

tan x * sin x + cos x = (sin x/cos x)sin x + cos x = sin^2 cos x + cos x = (1-cos^2x)/cos x + cos x = 1/cos x - cos x+ cos x = 1/cos x = sec x

NNADVOKAT [17]3 years ago
3 0

Answer:

sec(x)

Step-by-step explanation:

tan x * sinx + cos x

Tan x = sin x/ cos x

Replace tan x in the equation

sinx /cosx * sin x + cos x

sin^2 x / cos x + cos x

We know sin^2 x = 1 -cos^2 x

Replace sin^2 with 1 -cos ^2 x

(1 -cos^2 x) / cos x  + cos

Distribute the denominator

1 /cos x - cos x + cos x

Combine like terms

1/cos x

sec x

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If the number of defects found is more than one standard deviation above the mean, the manufacturing facility needs to recalibra
Sliva [168]

Answer:

16% probability that the facility needs to recalibrate their machines.

Step-by-step explanation:

We have to use the Empirical Rule to solve this problem.

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

What is the probability that the facility needs to recalibrate their machines?

They will have to recalibrate if the number of defects is more than one standard deviation above the mean.

We know that by the Empirical Rule, 68% of the measures are within 1 standard deviation of the mean. The other 100-68 = 32% is more than 1 standard deviation from the mean. Since the normal distribution is symmetric, of those 32%, 16% are more than one standard deviation below the mean, and 16% are more than one standard deviation above the mean.

So there is a 16% probability that the facility needs to recalibrate their machines.

7 0
4 years ago
If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes
Keith_Richards [23]
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
4 0
4 years ago
Read 2 more answers
Can someone help me?
Tresset [83]
Question 2 use 180 -115 and divide by 2

maybe
6 0
3 years ago
6(x+4)= how to solve this ?
tino4ka555 [31]

Answer: 6x + 24

Explanation: In this problem, the 6 "distributes" through the parenthses, which means that it multiplies by each of the terms inside.

So we have 6(x) + 6(4) which simplifies to 6x + 24.

7 0
3 years ago
Read 2 more answers
Which of the following explains why f(x) = log4x<br> does not have a y-intercept?
Margarita [4]

Answer: f(x) approaches infinity

Step-by-step explanation:

N/A

7 0
3 years ago
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