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qwelly [4]
3 years ago
10

Please help! The expressions 3x - 9 and 23 - 5x represent the lengths (in feet) of two sides of an equilateral triangle. Find th

e length of a side. Show your work.
Mathematics
1 answer:
valentina_108 [34]3 years ago
5 0
An equilateral triangle has 3 congruent sides so the 2 expressions above are equal
3x - 9= 23 - 5x
And then you solve for x
-22=-2x
X=11
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Lines a and b are cut by transversal f. At the intersection of lines f and a, the top left angle is 96 degrees. At the intersect
Sloan [31]

Answer: 22

Step-by-step explanation:

Lines a and b are cut by transversal f.

Then from the given information,

6x-36=96      [ If two parallel lines are cut by a transversal then, the alternate exterior angles are equal. ]

\Rightarrow\ 6x=96+36\\\\\Rightarrow\ 6x=132\\\\\Rightarrow\ x=\dfrac{132}{6}\\\\\Rightarrow\ x=22

Hence, the value of x so that lines a and b are parallel lines cut by transversal  = 22

5 0
3 years ago
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On the first month test Keiko answered 19 out of 25 questions correctly. On the second test she got 17 out of 20 correct. On whi
LekaFEV [45]

He/she idk, got a higher score on the second test. How you do it is divide the score by the number out of it so like 19 divided by 25. Umm for the ratio, I'm not really sure, I guess you just do 19:25 and 17:20

8 0
3 years ago
Which is the graph of arctan(x)?
Evgen [1.6K]

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The Inverse Tangent Function (arctan)

Graph of y = tan x.

6 0
3 years ago
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Please help and provide an explanation for why the answer is f(x) = 3/2x + 7
inna [77]

Answer:

see below

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the point-slope equation is

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2 years ago
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Select the correct expressions. 1 Identify each expression that represents the slope of a tangent to the curve y= * +1 at any po
Olegator [25]

The expressions which represents the slope of a tangent to the curve y=\frac{1}{x+1} at any point (x, y) are:

                                ​f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}

<h3>The slope of a tangent to the curve.</h3>

Mathematically, the slope of a tangent line to the curve is given by this equation:

f'(x) =limh \rightarrow 0\frac{f(x+h)-f(x)}{h}

Given the function:

f(x)=y=\frac{1}{x+1}

When (x + h), we have:

f(x+h)=y=\frac{1}{x+h+1}

Next, we would find the derivative of f(x):

f'(x) =limh \rightarrow 0\frac{\frac{1}{x+h+1} -\frac{1}{x+1}}{h}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -(x+h+1)}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{x+1 -x-h-1}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{h(x+1)(x+h+1)}\\\\f'(x) =limh \rightarrow 0\frac{-h}{x^2h+2xh+xh^2+h^2 +1}\\\\f'(x) = \frac{-1}{(x+1)(x+1)} \\\\f'(x) = \frac{-1}{(x+1)^2}

Read more on slope of a tangent here: brainly.com/question/26015157

#SPJ1

5 0
3 years ago
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