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slamgirl [31]
3 years ago
15

A jewelry store purchases necklaces for n dollars each. The necklaces are then marked up by 20%, and costumers are charged 5% sa

les tax on the selling price. The expression n+
Mathematics
1 answer:
Aleksandr [31]3 years ago
8 0

Answer:

Step-by-step explanation:

Cost price = $n

Mark up price = selling price- cost price

= (20% × $n)

Selling price = $n + (20% × $n)

Sales tax = 5% × ($n + (20% × $n))

Total cost = selling price + sales tax

Total cost = $n + (20% × $n) + 5% × $n + (20% × $n)

= 1.2 × $n + 0.06 × $n

= 1.26 × $n.

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RideAnS [48]

Answer:

3+g=- 134

Step-by-step explanation:

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3 years ago
Determine which equation is belongs to the graph of the limacon curve below.
JulsSmile [24]

Answer: Option C.

Step-by-step explanation:

We have functions of r(θ)

In our graph, we can see that the minimum value of r is when θ = 0°, and the maximum value is when θ = 180°.

We also know that the graph is in the square (-5, 5)x(-5, 5) and you can see that the maximum radius is almost less than 5.

Then let's analyze the options:

A) r = 3 - 2*cos(θ)

the maximum is at the right angle, but the maximum is:

r = 3 -2*(-1) = 5, so this maximum value is bigger than the one in the graph.

B) r = 3 - sin(θ)

For the sin functions, the maximum and minimum do not correspond with the values i write earlier, so we can discard this option.

C) r = 3 - cos(θ)

The maximum is: r = 3 - (-1) = 4, so this may be the correct answer.

the minimum is r = 3 - 1 = 2,

this is a possible equation for our circle.

D) r = 2 - 2*cos(θ)

Here, when θ = 0, we have: r = 2 - 2*1 = 0, but in the graph we can see that the radius is not 0 when θ = 0, so we can discard this option.

So the only option that makes sense is option C.

8 0
3 years ago
A) Find the lowest common multiple (LCM) of 36 and 48
dezoksy [38]
144 is the lowest hoped this helped
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3 years ago
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Find the missing angle.
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Step-by-step explanation:

3 0
3 years ago
Find the first three iterates of the function f(z) = z2 + c with a value of c = 2 - 3i and an initial value of z0 = 1 + 2i.
Artist 52 [7]

Answer:

C

Step-by-step explanation:

We have: (I rewrote the function)

f(z_n)={z_{n-1}} ^2+c

Given that:

\displaystyle c=2-3i \text{ and } z_0 = 1 + 2 i

The first iterate will be:

\displaystyle \begin{aligned} f(z_1)&=(z_0)^2+c \\ &=(1+2i)^2+(2-3i) \\ &= (1+4i+4i^2)+(2-3i) \\ &=1+4i-4+2-3i \\ &=-1+i \end{aligned}

The second iterate will be:

\begin{aligned}f(z_2) &=(z_1)^2+c\\ &=(-1+i)^2+(2-3i) \\&= (1-2i+i^2)+(2-3i) \\&=1-2i-1+2-3i \\&=2-5i \end{aligned}

And the third iterate will be:

\begin{aligned} f(z_3)&=(z_2)^2+c\\ &=(2-5i)^2+(2-3i) \\ &=(4-20i+25i^2)+(2-3i) \\ &=4-20i-25+2-3i \\ &=-19-23i \end{aligned}

Hence, our answer is C.

3 0
3 years ago
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