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3 years ago

Rotations Please help me.

Mathematics

1 answer:

Trava [24]3 years ago

7 0

Answer:

see attached

Step-by-step explanation:

You can do these yourself fairly easily. Copy the figure and the coordinate axes onto a piece of tracing paper (tissue paper or even facial tissue will work, too), then rotate the copied figure 90° clockwise.

Line up the origin of the axes and make sure the axes you drew line up with the ones on the original figure. For 90° clockwise rotation, what was the +y axis will now align with the +x axis. Copy the rotated figure from the tracing paper back to the graph on your problem page in its new location. (You can cut out the figure, if necessary. Just be sure to make note of the position relative to the axes.)

This sort of physical activity reinforces the thinking you need to do to mentally rotate the figures. It is worth the effort.

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What is a number that has 5 factors through 1-20

777dan777 [17]

16 beause 1x16 2x8 4x4 1 2 4 8 16

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3 years ago

Read 2 more answers

Calculus 2 Master needed, show steps with partial fraction decomposition <img src="https://tex.z-dn.net/?f=%5Cint%283x%5E2-26x%2

Vladimir79 [104]

Answer:

-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C

Step-by-step explanation:

The fraction will be split into a sum of two other fractions.

The first fraction will have a denominator of x − 5. The numerator will the a polynomial of one less order, in this case, a constant A.

The second fraction will have a denominator of x² + 4. The numerator will be Bx + C.

$\frac{3x^{2}-26x+26}{(x-5)(x^{2}+4)}=\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}$

Combine the two fractions back into one using the common denominator.

$\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}=\frac{A(x^{2}+4)+(Bx+C)(x-5)}{(x-5)(x^{2}+4)}$

This numerator will equal the original numerator.

$A(x^{2}+4)+(Bx+C)(x-5)=3x^{2}-26x+26\\Ax^{2}+4A+Bx^{2}-5Bx+Cx-5C=3x^{2}-26x+26\\(A+B)x^{2}+(C-5B)x+(4A-5C)=3x^{2}-26x+26$

Match the coefficients.

$A+B=3\\C-5B=-26\\4A-5C=26$

Solve the system of equations.

$A=-1\\B=4\\C=-6$

So we can rewrite the integral as:

$\int {(\frac{-1}{x-5}+\frac{4x-6}{x^{2}+4}) } \, dx$

Solving:

$\int {\frac{-1}{x-5}\, dx + \int {\frac{4x}{x^{2}+4}} \, dx - \int {\frac{6}{x^{2}+4} } \, dx$

$-\int {\frac{1}{x-5}\, dx + 2\int {\frac{2x}{x^{2}+4}} \, dx - 6\int {\frac{1}{x^{2}+4} } \, dx$

$-ln|x-5| + 2ln(x^{2}+4) - 6(\frac{1}{2} tan^{-1}(\frac{x}{2} )) + C$

$-ln|x-5| + 2ln(x^{2}+4) - 3 tan^{-1}(\frac{x}{2} ) + C$

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3 years ago

How many solutions does the following equation have?<br> 14x + 12] = 0

Marysya12 [62]

Answer:

$x=-\frac{6}{7}$

It has one solution

Step-by-step explanation:

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4 years ago

What is the prime factor of 8

ExtremeBDS [4]

Prime factor is a factor that is a prime number.
Prime factors of :
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${2}^{3}$
12 : 2 × 2 × 3. Also can be written as
${2}^{2} \times 3$

20 : 2 × 2 × 5. Also can be written as
${2}^{2} \times 5$

30 : 2 × 3 × 5

56 : 2 × 2 × 2 ×7. Also can be written as
${2}^{3} \times 7$

70 : 2 × 5 × 7

You can find all of them by using Factor Tree [ as shown in the picture] or Dividing by prime numbers.

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3 years ago

Sam rotated parallelogram ABCD 270 counter clockwise around the origin if angle a is 120 and angle B 60 what is the degree measu

jarptica [38.1K]

Given that ABCD is a parallelogram.

where angle B is 60 degree.

Notice that only parallelogram is rotated. Rotation doesn't changes the name of angles. So angle B will remain at it's position.

Hence value of angle B will also not change.

So final answer for the value of angle B is 60 degree.

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3 years ago