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zhuklara [117]
3 years ago
14

A _____ is a type of variable that is not accounted for in a research experiment.

Mathematics
2 answers:
Mariana [72]3 years ago
7 0
Dependent variable or confounding variable, hope this at least got u part of the way there try 2 search it up 
Phantasy [73]3 years ago
6 0
The Answer is C, The independent and the dependent variable are the x and y axis and the only one left is the confounding variable.

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6^2r=800 round it to the nearest hundreth
kogti [31]

Question: solve for r: 6 ^ (2r) = 800


Answer: 1.87


Step-by-step explanation:

6 ^ (2r) = 800

log(6 ^ (2r)) = log(800)

2r log(6) = log(800)

r = log(800) / (2 log(6))

r = 1.86537641979

round

r = 1.87

log can be any logarithm function.


6^(2×1.86) = 784.734328546

6^(2×1.87) = 813.365367336


7 0
3 years ago
Solve the System by Graphing:
Marat540 [252]

Answer:

  • B) One solution
  • The solution is  (2, -2)
  • The graph is below.

=========================================================

Explanation:

I used GeoGebra to graph the two lines. Desmos is another free tool you can use. There are other graphing calculators out there to choose from as well.

Once you have the two lines graphed, notice that they cross at (2, -2) which is where the solution is located. This point is on both lines, so it satisfies both equations simultaneously. There's only one such intersection point, so there's only one solution.

--------

To graph these equations by hand, plug in various x values to find corresponding y values. For instance, if you plugged in x = 0 into the first equation, then,

y = (-3/2)x+1

y = (-3/2)*0+1

y = 1

The point (0,1) is on the first line. The point (2,-2) is also on this line. Draw a straight line through the two points to finish that equation. The other equation is handled in a similar fashion.

7 0
2 years ago
2000/500 x47 + 50 -20
saul85 [17]
40 is the correct answer for this
6 0
3 years ago
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
What is the measure of angle TSU?
Elden [556K]

Answer:

63°

Step-by-step explanation:

TSU = 180 - 117 = 63

6 0
3 years ago
Read 2 more answers
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