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Studentka2010 [4]
3 years ago
9

If someone could help me with number 26...

Mathematics
1 answer:
Volgvan3 years ago
8 0
Given:
\text{sec x} -  \sqrt{\text{2 sec x}-1} =0

Solution:
To solve the equation, it would be best if we remove the root. We remove the root by squaring the equation, but first we need to move the root and the content to the left side.
\text{sec x} -  \sqrt{\text{2 sec x}-1} =0
\text{sec x} =  \sqrt{\text{2 sec x}-1}

Then square both side to remove the root
\text{sec}^2 x= \text{2 sec x}-1

After removing the root, move all terms to the left side
sec² x - 2 sec x + 1 = 0

Do factorization, remember that
a² - 2a + 1 = (a - 1)²
So,
sec² x - 2 sec x + 1 = 0
(sec x - 1)² = 0
sec x - 1 = 0
sec x = 1
\dfrac{1}{\text{cos x}} = 1
cos x = 1
cos x = cos 0°
x = 0°
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The ratio of boys to girls in a class is 2:3. there are twelve boys in the class. how many girls are there
gogolik [260]
The answer is 18

Solution: 2:3 2x6=12
3x6=18
Hope this help
4 0
3 years ago
Simplify the given expression. (16x^2 + 15xy – 19y^2) – (3x^2 – 3xy)
ruslelena [56]

Answer:

-19 y^2 + 18 x y + 13 x^2

Step-by-step explanation:

Simplify the following:

16 x^2 + 15 x y - 19 y^2 - (3 x^2 - 3 x y)

Factor 3 x out of 3 x^2 - 3 x y:

16 x^2 + 15 x y - 19 y^2 - 3 x (x - y)

-3 x (x - y) = 3 x y - 3 x^2:

16 x^2 + 15 x y - 19 y^2 + 3 x y - 3 x^2

Grouping like terms, 16 x^2 + 15 x y - 19 y^2 - 3 x^2 + 3 x y = -19 y^2 + (15 x y + 3 x y) + (16 x^2 - 3 x^2):

-19 y^2 + (15 x y + 3 x y) + (16 x^2 - 3 x^2)

x y 15 + x y 3 = 18 x y:

-19 y^2 + 18 x y + (16 x^2 - 3 x^2)

16 x^2 - 3 x^2 = 13 x^2:

Answer:  -19 y^2 + 18 x y + 13 x^2

4 0
3 years ago
Read 2 more answers
Which function grows
mojhsa [17]
Y = 8x + 29 is the answer for your question
4 0
3 years ago
A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
Find the correct linear equation: y is the quotient of thrice a number and ten.
Anarel [89]

Answer:

y = (3x)/10

Step-by-step explanation:

quotient  = divide

7 0
3 years ago
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