# Thread: An architect drawing !!!

1. ## An architect drawing !!!

i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem

i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem

The question as asked makes no sense. There must be at least one

RonL

i need help with some of those problems ,and please if anyone is going to solve them please give few tips on how did u solve the problem

Has this not already been posted?

RonL

4. 4. As $\displaystyle x^2 + x -1$ is positive between $\displaystyle x=1$ and $\displaystyle x=2$, the area is:

$\displaystyle A = \int_1^2 (x^2 + x -1) dx$

5. The first four terms in the expansion of $\displaystyle (1+x)(1-2x)^10$ may be found by expanding the second term using the binomial theorem and multiplying out:

$\displaystyle (1+x)(1-2x)^{10}=(1+x)[1+10(-2x)+\frac{10.9}{2!}(-2x)^2+\frac{10.9.8}{3!}(-2x)^3+..]$

and keeping the four terms with the lowest powers of $\displaystyle x$

RonL

5. ## thank u soo much for the answer

thanks alot for the answer, and by the way in number 2 there is nothing missing ,its just like that ...i myself am wondering ,anyways i would like to thank u for answering them for me and would like please to answer number 3 thanks the last one left in the home work . thank you , answer only if u can

thanks alot for the answer, and by the way in number 2 there is nothing missing ,its just like that ...i myself am wondering ,anyways i would like to thank u for answering them for me and would like please to answer number 3 thanks the last one left in the home work . thank you , answer only if u can
let's find the distance $\displaystyle PQ$ and $\displaystyle QM$ then, so we can find the area of the triangle.

to do that, we need to find points $\displaystyle P$ and $\displaystyle Q$

these points will be where the graphs intersect, that is, where

$\displaystyle x^2 - 2x + 2 = -x + 4$

$\displaystyle \Rightarrow x^2 - x - 2 = 0$

$\displaystyle \Rightarrow (x - 2)(x + 1) = 0$

$\displaystyle \Rightarrow x = 2 \mbox { or } x = -1$

when $\displaystyle x = 2$:

$\displaystyle y = -2 + 4 = 2$

when $\displaystyle x = -1$:

$\displaystyle y = 1 + 4 = 5$

so we have $\displaystyle P(-1,5)$ and $\displaystyle Q(2,2)$

Now, slope of line PQ is $\displaystyle \frac {y_2 - y_1}{x_2 - x_1} = \frac {3}{-3} = -1$

slope of line QM is $\displaystyle \frac {y_2 - y_1}{x_2 - x_1} = 1$

this means, line PQ is perpendicular to QM, and so $\displaystyle \triangle PQM$ is a right-triangle.

Now let's find the lengths of PQ and QM

By the distance formula:

$\displaystyle PQ = \sqrt {(2 - (-1))^2 + (2 - 5)^2} = \sqrt {18}$

$\displaystyle QM = \sqrt {(1 - 2)^2 + (1 - 2)^2} = \sqrt {2}$

$\displaystyle \Rightarrow A_{ \triangle PQM} = \frac {1}{2}bh = \frac {1}{2} \left( \sqrt {2} \right) \left( \sqrt {18} \right) = 3$

Now to find the area of the shaded region:

$\displaystyle A_{R} = \int_{-1}^{2} \left( -x + 4 - x^2 + 2x - 2 \right) dx$

$\displaystyle = \int_{-1}^{2} \left( -x^2 + x + 2 \right) dx$

$\displaystyle = \left[ - \frac {1}{3} x^3 + \frac {1}{2} x^2 + 2x \right]_{-1}^{2}$

$\displaystyle = 4 \frac {1}{2}$

Since $\displaystyle 4 \frac {1}{2} = 1 \frac {1}{2} \cdot 3$, we have the desired result