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7nadin3 [17]
3 years ago
14

HELPPPPPP PLSPLSPLSS A.50 B.85 C.60 D.75

Mathematics
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

D

Step-by-step explanation:

xeze [42]3 years ago
7 0

Answer:

it is 75 because I know it is 75

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As you can see on the number line, there is a closed dot on 4 and an open dot on 6.



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What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−23x+5y
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4 0
3 years ago
Read 2 more answers
How would you find the missing x and y values if you don’t have a model or equation to go by?
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3 0
3 years ago
I need help pls thank you and I'll send better pics
kakasveta [241]
\begin{gathered} c)\text{ output = (input}\times\text{ input) }+\text{ 1} \\ d)\text{ output = input + 5} \end{gathered}

Explanation:

\begin{gathered} c)\text{ Change in input = 4 - 3 , 3 - 2, 2 -1 } \\ \text{change in input = 1 (it is constant )} \\ \text{The output depends on the input.} \end{gathered}\begin{gathered} To\text{ get output 2: (1 }\times1\text{ )+ 1 = 1 +1 = 2} \\ To\text{ get output 5: (2}\times2)\text{ + 1 = 4 + 1 = 5} \\ To\text{ get output 10: (3}\times3)\text{ + 1 = 9 + 1 = 10} \\ To\text{ get output 17: (4}\times4)\text{ + 1 = 16 + 1 = 17} \\ To\text{ get output 26: (5}\times5)\text{ + 1 = 25 + 1 = 26} \end{gathered}\begin{gathered} \text{Formula for the Process = (input}\times\text{ input) }+\text{ 1} \\ Process=(input)^2\text{ }+\text{ 1} \\ \text{output = (input}\times\text{ input) }+\text{ 1} \end{gathered}

\begin{gathered} d)\text{ rate of change = }\frac{change\text{ in output }}{\text{change in input}} \\ \text{rate of change = }\frac{-5-(-15)}{0-(-2)}\text{ = }\frac{10-(-5)}{3-0}\text{ = }\frac{15-10}{4-3} \\ \text{rate of change = }\frac{10}{2}\text{ = }\frac{15}{3}=\frac{5}{1} \\ \text{rate of change = 5} \end{gathered}\text{output = input + 5}

5 0
1 year ago
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