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Marysya12 [62]
3 years ago
11

Graph the relation shown in the table. Is the relation a function? Why or why not?

Mathematics
2 answers:
ELEN [110]3 years ago
6 0

Answer:

the first one

Step-by-step explanation:

use the vertical line test. the horizontal line would intersect at two points.

dezoksy [38]3 years ago
4 0

Answer:

Not a function fails the vertical line test

Step-by-step explanation:

This is not a function.  The value of x = -1  goes to two different values of y

This would fail the vertical line test

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I need help plssssssssss
drek231 [11]

We can say that the area of the figure equals...

5.5 by 9 rectangle - 2 by 2 triangle - 2 by 5 triangle - 9 by 3.5 triangle

5.5 by 9 rectangle = 5.5 x 9 = 49.5 un²

2 by 2 triangle = 1/2 x 2 x 2 = 2 un²

2 by 5 triangle = 1/2 x 2 x 5 = 5 un²

9 by 3.5 triangle = 1/2 x 9 x 3.5 = 15.75 un²

area of figure = 49.5 un² - 2 un² - 5 un² - 15.75 un² = 26.75 un²

answer: 26.75 un²

5 0
3 years ago
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Determine the average rate of change of the function between the given values of the variable. g(x)=2/x ; x=4, x=a
IrinaK [193]
Average rate of change implies the quotient f(4)-f(a)/4-a which equals (1/2-a/2)/4-a=(1-a)/(8-2a) on (4,a)
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3 years ago
The zeroes are quadratic functions<br> are 6 and -4 which of these could be the functons.
LenaWriter [7]

Answer:

6 is the answer of your question

4 0
3 years ago
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Simplify the expression 5^3 x 5^-5
kotykmax [81]

Answer:

\frac{1}{5^2} --- 1 over 5 squared

Step-by-step explanation:

When multiplying terms with a common base, you just add the exponents:

x^a\times x^b=x^{a+ b}

That's true even when you don't have any exponents.

5\times5=5^1\times5^1=5^{1+1}=5^2=25

\rightarrow5^3\times5^{-5}\\\rightarrow5^{3-5}\\\rightarrow5^{-2}

A negative exponent isn't fully simplified, so there's another rule to use:

x^{-y}=\frac{1}{x^y}

That is '1 over x to the y' if it's too small to read.

\rightarrow5^{-2}=\frac{1}{5^2}

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2 years ago
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Here is all the work and answers for these questions they are all correct i checked them all and ive done this assignmnet before

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