Let the first number be = x
Then the second number = 2x
The third number = 2x - 5
Their sum = 55
This can be written in an equation as =
x + 2x + 2x - 5 = 55
= x + 2x + 2x = 55 + 5 ( transposing -5 from LHS to RHS changes -5 to +5 )
= x + 2x + 2x = 60
= 5x = 60
= x = 60 ÷ 5 ( transposing ×5 from LHS to RHS changes ×5 to ÷5 )
= x = 12
The first number = x = 12
The second number = 2x = 2 × 20 = 24
The third number = 2x - 5 = 24 - 5 = 19
Therefore , the three numbers are 12 , 24 and 19 .
<span>A(t) = −(t − 8)2 + 535
You would like to find the maximum of this function:</span> −(t − 8)^2<span>This part is always negative or zero as a number squared cannot be negative and you multiply by -1: Thus the maximum of this part MAX:</span>−(t − 8)^2=0
<span>The max will be when t=8 and its value is 535
</span>
This equation is written in <em>standard form</em>, so we need to change it into <em>slope-intercept form.</em>
<em />
3x = -y - 5
3x + y = -y + y - 5
3x + y = -5
3x - 3x + y = -5 - 3x
y = -5 - 3x
y = -3x - 5
The slope is classified as "m" in this type of equation, so, the slope is -3.
Best of Luck!
Answer:
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Step-by-step explanation:
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