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Andreas93 [3]
3 years ago
11

I need this answered plz!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

Red: Left side equation

Blue: Right side equation

2x+7=2(x-1)

2x+7=2x-2

2x-2x=-2-7

0x=-9= 0 No solution

Step-by-step explanation:

Use Desmos graphing calculator if u need help with graphs

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A man has 1 coins in his pocket, all of which are dimes and quarters. Aif the total value of his change is $2.75, how many dimes
VLD [36.1K]

Using a system of equations, it is found that there are 5 dimes and 9 quarters in his pocket.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are given as follows:

  • Variable x: number of dimes in his pocket.
  • Variable y: number of quarters in his pocket.

He has a total of 14 coins, hence:

x + y = 14 -> y = 14 - x.

They are worth $2.75, hence, considering the value of each coin(dimes $0.1 and quarters $0.25), we have that:

0.1x + 0.25y = 2.75

Since y = 14 - x:

0.1x + 0.25(14 - x) = 2.75

x = (0.25*14 - 2.75)/0.15.

x = 5.

y = 14 - x = 14 - 5 = 9.

There are 5 dimes and 9 quarters in his pocket.

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

4 0
2 years ago
at the end of his shift, Bruno had $340 worth of tips all in $10 and $20 bills .if he had two more $20 bills than $10 bill .how
PSYCHO15rus [73]
Bruno had 10 bills for each if you don't count the 2 more $20 bills. If you want a complete number, bruno had 12 $20 dollar bills and 10 $10 bills
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Find the Area and Perimeter of both figures!
Leokris [45]

where is the figures to solve.

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Here are the prices of five new skateboards, in dollars: 75, 82, 100, 120, 140.
Lynna [10]

The standard deviation<u> </u><u>INCREASES</u>

Step-by-step explanation:

Standard deviation is used to show how the points of the data deviate from the mean. The formulae for deriving standard deviation is attached. As seen from the formulae, the greater the variance of the data from the mean, the higher the Standard Deviation.

The mean of the given data points is $103.4. $450 is way off from this mean meaning that there is a large variance in this data point.

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3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

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

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
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