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jenyasd209 [6]
3 years ago
7

Nolan has $15.00 and he earns $6.00 an hour babysitting. The equation m = 6h + 15 can be used to determine how much money in dol

lars (m) Nolan has after any number of hours of babysitting ( h ).
If Nolan works for 40 hours this month, how much money will he have total?
Select one:
$285
$255
$240
$270
Mathematics
2 answers:
trasher [3.6K]3 years ago
6 0
Think about it 
Write a function rule that gives the total cost c(p) of p pounds of sugar if<span> each pound ... How </span>many <span>weeks </span>will<span> Melissa need to add to her savings before she </span>can<span> ... D {} 10 20 3{} </span>40<span> ${} .... </span>Nolan has $15.00<span>, and </span>he earns $6.00<span> an </span>hour babysitting<span>. ... </span>money<span> in </span>dollars<span> (</span>m<span>) </span>Nolan has after any number<span> of </span>hours<span> of </span>babysitting<span> (</span>h<span>).
</span>
SIZIF [17.4K]3 years ago
5 0
Answer is <span>$255

</span><span>m = 6h + 15
=6(40) + 15
=240 +15
=255</span>
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Intersection point of Y=logx and y=1/2log(x+1)
GalinKa [24]

Answer:

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

The Problem:

What is the intersection point of y=\log(x) and y=\frac{1}{2}\log(x+1)?

Step-by-step explanation:

To find the intersection of y=\log(x) and y=\frac{1}{2}\log(x+1), we will need to find when they have a common point; when their x and y are the same.

Let's start with setting the y's equal to find those x's for which the y's are the same.

\log(x)=\frac{1}{2}\log(x+1)

By power rule:

\log(x)=\log((x+1)^\frac{1}{2})

Since \log(u)=\log(v) implies u=v:

x=(x+1)^\frac{1}{2}

Squaring both sides to get rid of the fraction exponent:

x^2=x+1

This is a quadratic equation.

Subtract (x+1) on both sides:

x^2-(x+1)=0

x^2-x-1=0

Comparing this to ax^2+bx+c=0 we see the following:

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b=-1

c=-1

Let's plug them into the quadratic formula:

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}

x=\frac{1 \pm \sqrt{1+4}}{2}

x=\frac{1 \pm \sqrt{5}}{2}

So we have the solutions to the quadratic equation are:

x=\frac{1+\sqrt{5}}{2} or x=\frac{1-\sqrt{5}}{2}.

The second solution definitely gives at least one of the logarithm equation problems.

Example: \log(x) has problems when x \le 0 and so the second solution is a problem.

So the x where the equations intersect is at x=\frac{1+\sqrt{5}}{2}.

Let's find the y-coordinate.

You may use either equation.

I choose y=\log(x).

y=\log(\frac{1+\sqrt{5}}{2})

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

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3 years ago
Consider the polynomial operation 3x(x − 2)(5x + 2). Is the expression equivalent? Select Yes or No for A‐D.
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Answer:

see attached

Step-by-step explanation:

3x(x − 2)(5x + 2)

  • (x − 2)(5x + 2) expand
  • 5x² - 8x - 4
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3 0
3 years ago
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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8+4=10

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