we'll start off by grouping some

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)
<h2>Question 9:</h2>
1. Use Pythagorean Theorem (a²+b²=c²) to solve for missing side of triangle and rectangle. x²+16²=20², or x²+256=400. So, x²=144, and x=12
2. Use formula: 1/2(h)(b1+b2). 1/2 (12) (30+14).
3. Simplify: 1/2 (12) (44)=1/2(528)=264
Area of whole figure is 264 square mm.
<h2>Question 10:</h2>
Literally same thing but with trigonometry.
1. Use sine to find out length of dotted line: sin(60°)=x/12
2: Simplify: 12*sin(60°)=x. x≈10.4 (rounded to the nearest tenth)
3. Use Pythagorean Theorem to find out last leg of triangle: 10.4²+x²=12²
4: Simplify: 108.16 +x²=144. x²=35.84 ≈ 6
5: Use formula: 1/2(h)(b1+b2). 1/2 (10.4) (30+36)
6: Simplify: 1/2 (10.4) (66) =343.2
7: Area of figure is about 343.2
Remember, this is an approximate answer with rounding, so it might not be what your teacher wants. The best thing to do is do it yourself again.
3days=3chapters so he would have to read for 8 days
Answer:
x=6
Step-by-step explanation:
3х +5= 23
Subtract 5 from each side
3х +5-5= 23-5
3x = 18
Divide each side by 3
3x/3 =18/3
x = 6