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Svetradugi [14.3K]
3 years ago
15

8x + 2 – 6 = 4x +8 +3x

Mathematics
1 answer:
Viefleur [7K]3 years ago
4 0
The first step is to combine like terms. So u would do 2-6. You then get -4 then u just bring everything else down. So now u should have this :

8x -4 = 4x + 8 + 3x

Now you do the same thing to the other side. U combine like terms which in this case are 4x and 3x. So u just do 4+3 which is 7 so now u have 7x. So now ur equation would be:

8x -4 = 7x +8

Now u do the inverse of 7 which would be -7 and u do that on both sides. So it would look like this :

8x -4 = 7x + 8
-7x. -7x

So now 7 and -7 cancel each other other and 8 -7 is 1 so now u have one x. So ur equation would now look like this:

1x -4 = 8

Now u do the inverse of negative 4 which is +4. So you would add 4 on both sides on both sides. Like shown:

1x -4 = 8
+4 +4

So u just add 8 + 4 which is 12. Now since anything times 1 is itself ur answer will be X=12


Hope this helped



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Please help me idk this
PIT_PIT [208]

Answer:

SA=94

Step-by-step explanation:

SA= 2(4*5) + 2(4*3) + 2(5*3)

SA=2(20) + 2(12) + 2(15)

SA=40 + 24 + 30

SA=94

8 0
3 years ago
12345 what is the answer if you multiply 1x2x3x4x5 and do the same with subtraction, addition and devision, in that order. Then
Alexandra [31]

Well, we just need to perform the operations:

  • Multiplication: 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120
  • Subtraction: 1-2-3-4-5 = -13
  • Addition: 1+2+3+4+5 = 15
  • Division: 1 \div 2 \div 3 \div 4 \div 5 = \frac{1}{120}

So, if you add all the numbers together you get

120-13+15+\dfrac{1}{120} = 122 + \dfrac{1}{120}

Or, if you prefer,

\dfrac{1681}{120}

8 0
3 years ago
A patient is instructed to take three 50-mcg tablets of pergolide mesylate (Permax) daily. How many milligrams of the drug would
Black_prince [1.1K]

Answer:

The patient would receive 1.05mg of the drug weekly.

Step-by-step explanation:

First step: How many mcg of the drug would the patient receive daily?

The problem states that he takes three doses of 50-mcg a day. So

1 dose - 50mcg

3 doses - x mcg

x = 50*3

x = 150 mcg.

He takes 150mcg of the drug a day.

Second step: How many mcg of the drug would the patient receive weekly?

A week has 7 days. He takes 150mcg of the drug a day. So:

1 day - 150mcg

7 days - x mcg

x = 150*7

x = 1050mcg

He takes 1050mcg of the drug a week.

Final step: Conversion of 1050 mcg to mg

Each mg has 1000 mcg. How many mg are there in 1050 mcg? So

1mg - 1000 mcg

xmg - 1050mcg

1000x = 1050

x = \frac{1050}{1000}

x = 1.05mg

The patient would receive 1.05mg of the drug weekly.

6 0
3 years ago
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
What is 2^5 in standard notation
elena-14-01-66 [18.8K]
2x2x2x2x2 is 2^5 in standard notation
8 0
3 years ago
Read 2 more answers
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