Is 1/3 equivalent to 2/6?

Question 1

notsponge [240]

Answer:

6 means go to the right six times while -7 means go down seven times

Step-by-step explanation:

5 0

3 years ago

N over 7= 20<br> solve for N

Blababa [14]

Answer:

N = 140

Step-by-step explanation:

N/7 = 20

Inverse operation of division: multiplication

N/7 = 20

becomes:

N = 7 x 20

N = 140

Thus, the value of N is 140.

6 0

3 years ago

Line A goes through the points (1, 5) and (-1, 9). Line B goes through the points (0, 2) and (4, -6). Which of the following sta

Delvig [45]

B is correct. Explanation Below:
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Line A never intersects Line B, as they are parallel lines-the slopes are the same-and parallel lines never intersect.
Line A's Slope: (9-5)/(-1-1) = -2
Line B's Slope: (-6-2)/(4-0) = -2
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Hope this helps!

7 0

3 years ago

Compare and Contrast: Two equations are listed below. Solve each equation and compare the solutions. Choose the statement that i

Allisa [31]

<span>The correct answer is option B. i.e. Both equation 1 and 2 have the same number of solutions. The equation 1 is |5x+6| = 41 By removing the modulus sign we get, 5x + 6 = 41 and 5x +6 = -41. Solving, 5x = 6 = 41 we get, 5x = 35 or x = 7. And, on solving 5x +6 = -41 we get, 5x = -47 or x = 9.4 Now. the second equation is |2x+13| = 28 By removing the modulus sign we get, 2x+ 13 = 28 and 2x +13 = -28. Solving, 2x = 28 -13 = 15 or x = 7.5 . And, on solving 2x +13 = -28 we get, 2x = -41 or x = -20.5.TH </span>

4 0

3 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

4 years ago