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statuscvo [17]
4 years ago
9

30 Points! Please help! Is this a function? (-8,-16), (-2,-8), (-2,0), (6,4), (8,12)

Mathematics
1 answer:
levacccp [35]4 years ago
4 0
No this is not a function because it is not one to one! Each x value needs to have only one output! The -2 values have two different outputs thus making it not a function.
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PLEASE HELP ME AGAIN!!! :>
Nadusha1986 [10]

Answer: I think it's either C or A but, my gut is saying A :)

8 0
3 years ago
In a 45-45 right triangle, the two legs have the same length. in the figure(figure 4) both are given lengths of 1 . what is the
motikmotik
The hypotenuse formula is a^2+b^2=c^2
so each side a=1
                     b=1
                         1^2 +1^2=c^2
                          1+1= c^2   so the hypotenuse equals 2
7 0
3 years ago
Evaluate the expression 4.02 + 3y for x = 5 and y = 6​
IgorC [24]

the expression looks like its missing the x value tbh...

you provided x and y but the expression doesn't

have an x.

so, that applied, 4.02+3(6)

4.02+3(6)

4.02+18

22.02

------------------

but.... if you wrote your expression wrong and you meant to write 4.02x+3y,

than:

4.02(5) + 3(6)

20.1+18

= 38.1

__________

So, if you actually meant 4.02 + 3y, ur answer would be 22.02, but if you meant 4.02x+3y, then your answer would be 38.1

6 0
3 years ago
How many 3’s are in 1/8?
Murrr4er [49]

Answer:

There are none

Step-by-step explanation:

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Just divide 1 over 8

8 0
4 years ago
What is the zero of y=-4x^2+8x-28
tangare [24]

Solution, y-4x^2+8x-28=0:\quad \mathrm{Parabola\:with\:vertex\:at}\:\left(h,\:k\right)=\left(1,\:24\right),\:\mathrm{and\:focal\:length}\:|p|=\frac{1}{16}

Steps:

\mathrm{Rewrite}\:y-4x^2+8x-28=0\:\mathrm{in\:the\:standard\:form}

\mathrm{Rewrite\:as}, y=4x^2-8x+28

\mathrm{Complete\:the\:square}\:4x^2-8x+28:\quad 4\left(x-1\right)^2+24, y=4\left(x-1\right)^2+24

\mathrm{Subtract\:}24\mathrm{\:from\:both\:sides}, y-24=4\left(x-1\right)^2

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4, \frac{1}{4}\left(y-24\right)=\left(x-1\right)^2

\mathrm{Rewrite\:in\:standard\:form}, 4\cdot \frac{1}{16}\left(y-24\right)=\left(x-1\right)^2

\mathrm{Therefore\:parabola\:properties\:are:} \left(h,\:k\right)=\left(1,\:24\right),\:p=\frac{1}{16}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-\:austint1414}

8 0
3 years ago
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