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4 years ago

PLEASE HELP (algebra 1)

1 answer:

6 0

A) 6(4)^1-6(4)^0 = 24-6 = 18
6(4)^3-6(4)^2 = 384-96 = 288
b) It is 27.11 times difference. This is because g(x) is an exponential function and so as x increases, g(x) increases faster.

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How do you answer the equation -4r+11=4(1-r)+7?

kow [346]

Answer:

Any value of

makes the equation true.

All real numbers

Interval Notation:

(-Any value of

makes the equation true.

All real numbers

Interval Notation:

( ∞ ,∞ )

Step-by-step explanation:

3 0

3 years ago

At an effective annual interest rate of i > 0, each of the following two sets of payments has present value K: (i) A payment

IgorLugansk [536]

Answer:

The present value of K is, $K=251.35$

Step-by-step explanation:

First of all, we need to construct an equation system, so

$(1)K=\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}$

$(2)K=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

Then we equalize both of them so we can find $i$

$(3)\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

To solve it we can multiply $(3)*(1+i)^{4}$ to obtain $(1+i)^{4}*(\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}})$ , then we have $225(1+i)^{2}+225=169(1+i)^{3}+169(1+i)^{2}$ .

This leads to a third-grade polynomial $169i^{3}+451i^{2}+395i-112=0$ , after computing this expression, we find only one real root $i=0.2224$ .

Finally, we replace it in (1) or (2), let's do it in (1) $K=\frac{169}{(1+0.2224)} +\frac{169}{(1+0.2224)^{2}}\\\\K=251.35$

8 0

3 years ago

In order to give a souvenir table tennis ball to each participant in a youth table tennis tournament, a city recreation departme

professor190 [17]

6 boxes (add the cumulatively)
each box bas 5 pkgs so 10w,5o
1st box: 10w,5o
2nd box:20w,10o
3rd box:30w,15o
4th box:40w,20o
5th box:50w,25o
6th box:60w,30o
You need a min. of 48w and 27o

8 0

3 years ago

The cost, C, to produce b baseball bats per day is modeled by the function C(b) = 0.06b2 – 7.2b + 390. What number of bats shoul

kozerog [31]

Check the picture below, that's just an example of a parabola opening upwards.

so the cost equation C(b), which is a quadratic with a positive leading term's coefficient, has the graph of a parabola like the one in the picture, so the cost goes down and down and down, reaches the vertex or namely the minimum, and then goes back up.

bearing in mind that the quantity will be on the x-axis and the cost amount is over the y-axis, what are the coordinates of the vertex of this parabola? namely, at what cost for how many bats?

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ C(b) = \stackrel{\stackrel{a}{\downarrow }}{0.06}b^2\stackrel{\stackrel{b}{\downarrow }}{-7.2}b\stackrel{\stackrel{c}{\downarrow }}{+390} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\bf \left( -\cfrac{-7.2}{2(0.06)}~~,~~390-\cfrac{(-7.2)^2}{4(0.06)} \right)\implies (60~~,~~390-216) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\textit{number of bats}}{60}~~,~~\stackrel{\textit{total cost}}{174})~\hfill$