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anyanavicka [17]

3 years ago

7

MATH!!!!!!!!!!! OVER HERE / At the beginning of March, a store bought a Halloween mask at a cost of $10 and marked it up 60%. At

the end of the month, the Halloween mask had not sold, so the store marked it down 40%. What was the discounted price?

1 answer:

STatiana [176]3 years ago

6 0

Answer:

9.60

Step-by-step explanation:

100%->10

160%->16

-------------------------------------

100%->16

100%-40%=60%

60%->9.60

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attashe74 [19]

Answer:

Is it 1,512?

Step-by-step explanation:

6 0

3 years ago

22. An employee joined a company in 2017 with a starting salary of $50,000. Every year this employee receives a raise of $1000 p

Setler79 [48]

Answer:

(a) The recurrence relation for the salary is

$S_{n+1}=1.05*S_n+1000\\\\S_0=50000$

(b) The salary 25 years after 2017 will be $217044.85.

(c) $S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n$

Step-by-step explanation:

We can define the next year salary $S_{n+1}$ as

$S_{n+1}=S_n+1000+0.05*S_n=1.05*S_n+1000$

wit S0=$50000

If we extend this to 2 years from 2017 (n+2), we have

$S_{n+2}=1.05*S_{n+1}+1000=1.05*(1.05*S_n+1000)+1000\\S_{n+2} =1.05^2*S_n+1.05*1000+1000\\S_{n+2}=1.05^2*S_n+1000*(1.05^1+1)$

Extending to 3 years (n+3)

$S_{n+3}=1.05*S_{n+2}+1000=1.05(1.05^2*S_n+1000*(1.05^1+1))+1000\\\\S_{n+3}=1.05^3S_n+1.05*1000*(1.05^1+1)+1000\\\\S_{n+3}=1.05^3*S_n+1000*(1.05^2+1.05^1+1)$

Extending to 4 years (n+4)

$S_{n+4}=1.05*S_{n+3}+1000=1.05*(1.05^3*S_n+1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1.05*1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1000*(1.05^3+1.05^2+1.05^1+1.05^0)$

We can now express a general equation for S_n (salary at n years from 2017)

$S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n$

The salary at 25 years from 2017 (n=25) will be

$S_{25}=1.05^{25}S_0+1000*\sum_{0}^{24}1.05^i\\\\S_{25}=3.386*50000+1000*47.72=217044.85$

8 0

4 years ago

The price for 7 lb of potatoes is $5 what is the price per pound.

Thepotemich [5.8K]

Answer:

The answer is $1.4.

6 0

3 years ago

Which equation is equivalent to 16 Superscript 2 p Baseline = 32 Superscript p 3?.

Mrac [35]

The equation which is equivalent to 16 Superscript 2 p Baseline equal to 32 Superscript p 3 is,

$2^{8p}=2^{5p+15}$

<h3>What is equivalent equation?</h3>

Equivalent equation are the expression whose result is equal to the original expression, but the way of representation is different.

Given information-

The given equation in the problem is,

$16^{2p}=32^{p+3}$

Write both the equation in the form of same base number as,

$(2^4)^{2p}=(2^5)^{p+3}$

The power of the power of a number can be written as product of both the numbers. Thus,

$(2)^{4\times2p}=(2)^{5\times(p+3)}\\2^{8P}=2^{5P+15}$

This is the required equation.

Now if the base is the same at both side of the expression, then the powers can be compared. Thus,

$8p=5p+15$

Solve it further to find the value of p as,

$8p-5p=15\\3p=15\\p=5$

Thus the equation which is equivalent to 16 Superscript 2 p Baseline equal to 32 Superscript p 3 is,

$2^{8p}=2^{5p+15}$

Learn more about the equivalent expression here;

brainly.com/question/2972832

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2 years ago

The total surface area of the triangular prism is blank squar millimeter.

Shtirlitz [24]

Answer:

To find the area of a triangle, multiply the base by the height, and then divide by 2. The division by 2 comes from the fact that a parallelogram can be divided into 2 triangles. For example, in the diagram to the left, the area of each triangle is equal to one-half the area of the parallelogram

Step-by-step explanation:

hope this helps :)

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3 years ago