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3 years ago

Which best describes the polygon whose vertices in the coordinate plane are (-2,3),(2,3),(2,-1),(-2,-1)

1 answer:

7 0

The polygon is a perfect square. I graphed those points on a graph. I hope this helped if not let me know so I can try and write the correct answer.

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Are the triangles similar help.

xxTIMURxx [149]

Answer:

yes the two triangles are similar

7 0

3 years ago

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Find an equation of a plane that satisfies the given conditions

guajiro [1.7K]

The conditions are needed to be able to answer the question

3 0

3 years ago

What is 1/2(x+14) because i am having trouble solving it and no calculator has been able to help me also.

ivann1987 [24]

Hi there!

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I believe your answer is:

$\frac{1}{2}x+7$

»»————-　★　————-««

Here’s why:

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$\boxed{\text{Simplifying the equation...}}\\\\\frac{1}{2}(x+14)\\------------\\\rightarrow \frac{1}{2}*x = \frac{1}{2}x\\\\\rightarrow\frac{1}{2}*14=7\\\\\rightarrow \frac{1}{2}(x+14)= \boxed{\frac{1}{2}x+7}$

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»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.

7 0

3 years ago

Make B the subject of the relation : 1/a is equal to 1/b + c

stealth61 [152]

Answer:

By making ‘a’ the subject, I believe you mean isolate the variable ‘a’.

1/a - 1/b = 1/c : add 1/b to both sides

1/a = 1/b + 1/c : combine the unlike fractions by finding a common denominator, bc is the common denominator

1/a = (1/b)(c/c) + (1/c)(b/b) : simplify

1/a = (c/bc) + (b/bc) : add numerators only, because the denominators match

1/a = (c + b)/bc : multiply both sides by a

1 = (a)[(c + b)/bc] : multiply both sides by the reciprocal of [(c + b)/bc] which is [bc/(b + c)]

1[bc/(b + c)] = a

a = bc/(b + c)

This will not work if c = -b, because then you would be dividing by zero.

Example: 1/2 - 1/3 = 1/6 a = 2, b = 3 c= 6

a = bc/(b + c) => 2 = (3 x 6)/(3 + 6) => 2 = 18/9 => 2 = 2.

Step-by-step explanation:

4 0

4 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