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lukranit [14]
3 years ago
15

In the figure below, m 2 = 148º. Find 1​

Mathematics
1 answer:
omeli [17]3 years ago
8 0

Answer:

1    =      32

Step-by-step explanation:

the total is 180 so 180 - 148 = 32

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Which of the following expressions is equal to x² + 36? *
mash [69]

Answer:

{x}^{2}  + 36 \\ factors : -  6i \: and \: 6 i\\   = {x}^{2}  + 6i - 6i + 36 \\  = (x - 6i)(x - 6i) \\  = ( {x - 6i)}^{2}

8 0
3 years ago
F(x) = 2x + 6, g(x) = 4x2 Find (f + g)(x).
son4ous [18]
(f+g)(x) is simply saying add these two problems together.

(2x+6) + (4x^2) 

Parenthesis are now useless, as there is only addition.

2x + 6 + 4x^2

Reorder, cause it's pretty and familiar

4x^2 + 2x + 6

We could further simplify

2(2x^2 + x + 3)
5 0
4 years ago
Give the null and alternative hypotheses in symbolic form that would be used in a hypothesis test of the following claim:
irakobra [83]

Answer:

Step-by-step explanation:

The null hypothesis is usually the default statement. The alternative is the opposite of the null and usually tested against the null hypothesis

In this case study,

The null hypothesis in would be that the mean time between clicks of the second hand on a particular clock is 1 second. In symbolic form it would be u = 1

The alternative hypothesis would be that the mean time between clicks of the second hand on a particular clock is 1 not second. In symbolic form, it would be: u =/ 1

3 0
4 years ago
FIRST PERSON TO GET CORRECT GETS BRAINLIEST!!!!!!!!1
schepotkina [342]

Answer:

Option C. (-x,y)

Step-by-step explanation:

we know that

A point reflected across the y-axis has the following rule

(a,b)------> (-a,b)

so

If the center of the original circle was located at (x, y)

then

the center of the new circle after the transformation will be (-x,y)

3 0
3 years ago
Read 2 more answers
PLEASE HELP
kogti [31]
\textbf{You have to} restrict the domains of quadratic functions and absolute value functions, because these functions are \textbf{many-to-one} functions. For instance, the quadratic function f(x) = x^2 pairs both −2 and 2 with 4, and the absolute value function f(x) = |x| pairs both −2 and 2 with 2.

Linear functions (excluding constant functions) and exponential functions are \textbf{one-to-one} functions, so their domains \textbf{do not need} to be restricted.
_________________

An absolute value function, without domain restriction, has an inverse that is NOT a function.

In order to guarantee that the inverse must also be a function, we need to restrict the domain of the absolute value function to make it a one-to-one function.
7 0
4 years ago
Read 2 more answers
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