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RSB [31]
3 years ago
8

What is the value of k in the equation k/3 - 5 = 34? 11 1/3 13 102 117

Mathematics
2 answers:
mestny [16]3 years ago
6 0

1. Add 5 to both sides

k/3 = 34 + 5

2. Simplify 34 + 5 to 39

k/3 = 39

3. Multiply both side by 3

k = 39 * 3

4. Simplify 39 * 3 to 117

k = 117

Mamont248 [21]3 years ago
4 0

Answer:

D is the answer on Edgen.

Step-by-step explanation:

I took the test.

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Someone please help me with this​
Sedbober [7]

Answer:

The area of the shaded figure is:

  • <u>20 units^2</u>

Step-by-step explanation:

To obtain the area of the shaded figure, first, you must calculate this as a rectangle, with the measurements: wide (4 units), and long (6 units):

  • Area of a rectangle = long * wide
  • Area of a rectangle = 6 * 4
  • Area of a rectangle = 24 units^2

How the figure isn't a rectangle, you must subtract the triangle on the top, so, now we calculate the area of that triangle with measurements: wide (4 units), and height (2 units):

  • Area of a triangle = \frac{wide*height}{2}
  • Area of a triangle = \frac{4*2}{2}
  • Area of a triangle =\frac{8}{2}
  • Area of a triangle = 4 units^2

In the end, you subtract the area of the triangle to the area of the rectangle, to obtain the area of the shaded figure:

  • Area of the shaded figure = Area of the rectangle - Area of the triangle
  • Area of the shaded figure = 24 units^2 - 4 units^2
  • <u>Area of the shaded figure = 20 units^2</u>

I use the name "units" because the exercise doesn't say if they are feet, inches, or another, but you can replace this in case you need it.

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What is the result of 2 divided by StartFraction 4 Over 10 EndFraction? A number line going from 0 to 2. One-fifth Four-fifths 5
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Read 2 more answers
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
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