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Fudgin [204]
4 years ago
7

Can I have help answering these three problems?

Mathematics
1 answer:
Vsevolod [243]4 years ago
8 0

Problem One is B. y= -1/7 - 5/2

Problem Two is C. y= 0.2x + 3.4

Problem Three is A. no correct answer is given.

I hope this helps!

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After being discounted 40%, a weather radio sells for $41.97. Find the original price. (Round your answer to the nearest cent.)
o-na [289]
40%= $41.97 + 40%=$41.97 = 83.94
that is 80%now to find the 20% i dividied the 40% which is $20.98 so then I added the 80% and 20% which is $104.92

so the original price is $104.92
40%=83.94
+
40%=83.94
+
20%=20.98
=104.92
3 0
3 years ago
Kathi and Robert Hawn had a pottery stand at the annual Skippack Craft Fair. They sold some of their pottery at the original pri
Vinil7 [7]

Answer:

The number sold at $12.50 is  25

The number sold at $9.50 is 62

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations:

x 12.50 + y 9.50=901.50

x+y= 87

Where

x= number of pottery pieces sold

y= number of pottery pieces sold at a decreased price  

Multiplying the second equation by 9.50, and subtracting the second equation to the first equation:

12.50 x + 9.50y = 901.50

-

9.50 x+ 9.50 y= 826.5

__________________

3 x = 75

Solving for x

x =75/3

x = 25

Replacing x=25 on any equation:

x+y= 87

25+y =87

y=87-25

y= 62

5 0
4 years ago
Ik it's prob easy ques but help me i forgot everything
MaRussiya [10]

Answer:

The answer is 14

Step-by-step explanation:

I used a calculator because when you grow older and have a phone, logically you will always have a calculator :)

5 0
4 years ago
Read 2 more answers
100
Yuliya22 [10]
HznxbxjzjsjxjxjjJsjcjxjjajajxx
3 0
3 years ago
About Prove each of the following statements using mathematical induction. (a) Define the sequence {cn} as follows: c0 = 5 ck =
strojnjashka [21]

a.

\begin{cases}c_0=5\\c_n={c_{n-1}}^2&\text{for }n\ge1\end{cases}

For n=1, we have by this definition c_1={c_0}^2=5^2 which is equal to 5^{2^n} for n=1.

Assume c_n=5^{2^n} for some n=k. We want to use this to show that this implies c_{k+1}=5^{2^{k+1}}. By the recursive definition,

c_{k+1}={c_k}^2=(5^{2^k})^2=5^{2^{k+1}}

and we're done.

b.

\begin{cases}b_0=1\\b_n=2b_{n-1}+1&\text{for }n\ge1\end{cases}

For n=0, we have b_0=1, and 2^{n+1}-1=1.

Assume b_n=2^{n+1}-1 for some n=k. Then for n=k+1, we have

b_{k+1}=2b_k+1=2\left(2^{k+1}-1\right)+1=2^{k+2}-1

and we're done.

8 0
3 years ago
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