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Alex Ar [27]
3 years ago
8

I need help please and thank you!!!!

Mathematics
1 answer:
Naddika [18.5K]3 years ago
4 0

Answer:

Bottom left graph.

Step-by-step explanation:

I answered this by finding the y-int.

8(0) - 3y = 18

-3y = 18

y = -6

And it has a positive relationship so the line slants from bottom left to top right.

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Komok [63]

Answer:

C, finding temperature in June

6 0
3 years ago
Find the range of the function<br> F(x) = (x-2)2 +2
viktelen [127]
(x - 2)^2 will always be positive and will have a minimum value of  0 

so f(x) will have minimum of 2

Range is [2,∞)
8 0
3 years ago
I REALLY NEED HELP WORTH 10 POINTS URGENT!!!!!
polet [3.4K]

Answer:

(f\circ g)(4)=31

Step-by-step explanation:

<u>Composite Function</u>

Given f(x) and g(x) real functions, the composite function named fog(x) is defined as:

(f\circ g)(x)=f(g(x))

For practical purposes, it can be found by substituting g into f.

We have:

f(x)=3x+1

g(x)=x^2-6

Computing the composite function:

(f\circ g)(x)=f(g(x))=3(x^2-6)+1

Operating:

(f\circ g)(x)=3x^2-18+1

Operating:

(f\circ g)(x)=3x^2-17

Now evaluate for x=4

(f\circ g)(4)=3(4)^2-17=48-17

\boxed{(f\circ g)(4)=31}

8 0
3 years ago
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galina1969 [7]
Checitgimytgg 4567 friend
8 0
3 years ago
AB is tangent to O. If AO = 32 and BC = 98, what is AB? my guess is 130? please help!
Juli2301 [7.4K]
    
CO = AO  = 32    (Are rays.)
BO = BC + CO = 98 + 32 = 130
AB ⊥ BO
⇒  ΔAOB   is a rectangular triangle.
We use Pitagora's theorem.


\displaystyle\\&#10;AB =  \sqrt{BO^2 - AO^2}  = \sqrt{130^2 -32^2}=\\\\&#10;= \sqrt{16900 -1024}= \sqrt{15876}= \boxed{126 }



7 0
4 years ago
Read 2 more answers
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