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2 years ago

If mABC= 184°, what is m∠ABC?88°90°84°92°

Mathematics

1 answer:

jasenka [17]2 years ago

6 0

mADC=360-mABC=360-184=176⁰

So, measure of angle ABC= mADC/2=176/2=88⁰

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A credit card issuer charges an APR of 13.99%, and its billing cycle is 30 days long. What is its periodic interest rate?

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A credit card issuer charges an APR of 13.99%, and its billing cycle is 30 days long its periodic interest rate would be 1.15%. The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.

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which of the sets shown includes the elements of set Z that both negative numbers and multiples of ?. Z={-34,-28,-16,-2,4,8,12,2

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Answer:

multiples of 2

Step-by-step explanation:

6 0

2 years ago

F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

34kurt

Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

f(x) and g(x) and not inverse functions

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

$f(x) = 3x^2 + 9$

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

Hence, f(x) and g(x) and not inverse functions

4 0

2 years ago

A textbook store sold a combined total 402 of psychology and math textbooks in a week. The number of psychology textbooks sold w

uranmaximum [27]

Answer:

The number of textbooks of each type were sold is 134 math and 268 psychology books.

Step-by-step explanation:

Given:

Total number of math and psychology textbooks sold in a week is 402.

Now, let the number of math textbooks sold be $x$ .

And, the number of psychology textbooks be $2x$ .

According to question:

$x+2x=402$

$3x=402$

Dividing both sides by 3 we get:

$x=134$

So, total number of math textbooks were 134 .

And, total number of psychology textbooks were $2x=2\times 134$

$=268$ .

Therefore, the number of textbooks of each type were sold is 134 math and 268 psychology books.

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2 years ago

Below is the five Number summary for a 136 hikers who recently completed the John muir trail JMT the variable is the amount of t

Helga [31]

Answer:

$IQR = Q_3 -Q_1 = 28-18 = 10$

Now we can find the limits in order to determine outliers like this:

$Left = Q_1 -1.5 IQR = 18 -1.5*10=3$

$Right = Q_1 +1.5 IQR = 18 +1.5*10=33$

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

Step-by-step explanation:

For this case we have the following summary:

$Minimum = 9$ represent the minimum value

$Q_1 = 18$ represent the first quartile

$Median =Q_2= 21$ represent the median

$Q_3 = 28$ represent the third quartil

$Maximum=56$ represent the maximum

If we use the 1.5 IQR we need to find first the interquartile range defined as:

$IQR = Q_3 -Q_1 = 28-18 = 10$

Now we can find the limits in order to determine outliers like this:

$Left = Q_1 -1.5 IQR = 18 -1.5*10=3$

$Right = Q_1 +1.5 IQR = 18 +1.5*10=33$

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

7 0

2 years ago