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mixas84 [53]
3 years ago
6

2(x-3)= 2x +5 help me out please

Mathematics
2 answers:
Igoryamba3 years ago
8 0

Answer:

Look Down Low

Step-by-step explanation:

2(x - 3)  =  2x + 5

Multiply out the left side:

2x - 6  =  2x + 5

Subtract 2x from both sides:

   -6  =  5

Since this is impossible, this "equation" cannot be solved.

GalinKa [24]3 years ago
8 0

Answer:

11

Step-by-step explanation:

2(x - 3) = 2x + 5

Distribute 2 into (x - 3) on the left

2 X x -2 X 3 = 2x + 5

2x - 6 = 2x + 5

Add 6 to both sides to eliminate 6 on the left

2x - 6 + 6 = 2x + 5 + 6

2x = 2x + 11

Subtract 2x from both sides to put x on one side

2x - 2x = 2x - 2x + 11

0 = 11

11

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HELP ME HELP ME 10 POINTS
wel

Answer:

N+P or number 3

Step-by-step explanation:

7 0
3 years ago
A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of
GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.989.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

8 0
3 years ago
Here take these points and have fun with it
deff fn [24]

Answer:

thxs :)

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
adoni [48]

Answer: The required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

Step-by-step explanation:

Since we have given that

y=\ln[x(2x+3)^2]

Differentiating log function w.r.t. x, we get that

\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}

Hence, the required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

3 0
3 years ago
Kristen and Mitchell were given the same math assignment. Kristen completes 0.8 of her work in class. Mitchell completes 0.75 of
monitta
Mitchell because if you add a zero at the end of the .8 it would be 0.80 and 0.75 is smaller than 0.80 so Mitchell has more work to do at home
5 0
3 years ago
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