Answer:
m<Q = 92 degrees
Step-by-step explanation:
Opposite angles of a quadrilateral inscribed in a circle are supplementary.
Angles O and Q are opposite angles, so they are supplementary, so their measures add up to 180 degrees.
m<O + m<Q = 180
2x + 2x + 4 = 180
4x + 4 = 180
4x = 176
x = 44
m<Q = 2x + 4 = 2(44) + 4 = 92
We use quadratic formula for that one
remember that √-1=i
so
for
ax^2+bx+c=0
![x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%2B%2F-%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
so
given
1x^2+5x-5=0
a=1
b=5
c=-5
![x=\frac{-5+/-\sqrt{5^2-4(1)(-5)}}{2(1)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-5%2B%2F-%5Csqrt%7B5%5E2-4%281%29%28-5%29%7D%7D%7B2%281%29%7D)
![x=\frac{-5+/-\sqrt{25+20}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-5%2B%2F-%5Csqrt%7B25%2B20%7D%7D%7B2%7D)
![x=\frac{-5+/-\sqrt{45}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-5%2B%2F-%5Csqrt%7B45%7D%7D%7B2%7D)
![x=\frac{-5+/-3\sqrt{5}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-5%2B%2F-3%5Csqrt%7B5%7D%7D%7B2%7D)
solutions:
Answer:
3(x-25)
Step-by-step explanation:
Answer:
10 pounds of trial mix that costs $2.45 and 20 pounds of trial mix that costs $2.30
Step-by-step explanation:
![\\\text{Let x pounds of the trial mix that costs \$2.45 per pound, and y pounds of the trial mix}\\\text{that costs \$2.30 per pound are added to get a 30 pounds of trial mix }\\\text{that costs \$2.35 per pound. since the total mixture is 30 pounds, so we get}\\\\x+y=30\ \ \ \ \ \ \ \ \ \ \ \ \ ...... (i)\\\\\text{and the total cost equation is given by}\\\\\$2.45 x + \$2.30 y=\$2.35 (30)\\\\2.45 x + 2.30 y=70.5 \ \ \ \ \ \ \ \ \ \ \ \ \ ...... (ii)\\](https://tex.z-dn.net/?f=%5C%5C%5Ctext%7BLet%20x%20pounds%20of%20the%20trial%20mix%20that%20costs%20%5C%242.45%20per%20pound%2C%20and%20y%20pounds%20of%20the%20trial%20mix%7D%5C%5C%5Ctext%7Bthat%20costs%20%5C%242.30%20per%20pound%20are%20added%20to%20get%20a%2030%20pounds%20of%20trial%20mix%20%7D%5C%5C%5Ctext%7Bthat%20costs%20%5C%242.35%20per%20pound.%20since%20the%20total%20mixture%20is%2030%20pounds%2C%20so%20we%20get%7D%5C%5C%5C%5Cx%2By%3D30%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%20......%20%28i%29%5C%5C%5C%5C%5Ctext%7Band%20the%20total%20cost%20equation%20is%20given%20by%7D%5C%5C%5C%5C%5C%242.45%20x%20%2B%20%5C%242.30%20y%3D%5C%242.35%20%2830%29%5C%5C%5C%5C2.45%20x%20%2B%202.30%20y%3D70.5%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20......%20%28ii%29%5C%5C)
![\text{now to solve the equation (i) and (ii), we'll substitute }y=30-x \text{ from (i) into (ii)}\\\text{so we get}\\\\2.45x+2.30(30-x)=70.5\\\\2.45x+69-2.30x=70.5\\\\\Rightarrow 2.45x-2.30x=70.5-69\\\\\Rightarrow 0.15x=1.5\\\\\Rightarrow x=\frac{1.5}{0.15}\\\\\Rightarrow x=10\\\\\text{plug this value of x in (i), we get}\\\\y=30-x=30-10=20\\\\\text{so we must add 10 Pounds of trial mix that costs \$ 2.45 and 20 pounds of}\\\text{the trial mix that costs \$2.30}](https://tex.z-dn.net/?f=%5Ctext%7Bnow%20to%20solve%20the%20equation%20%28i%29%20and%20%28ii%29%2C%20we%27ll%20substitute%20%7Dy%3D30-x%20%5Ctext%7B%20from%20%28i%29%20into%20%28ii%29%7D%5C%5C%5Ctext%7Bso%20we%20get%7D%5C%5C%5C%5C2.45x%2B2.30%2830-x%29%3D70.5%5C%5C%5C%5C2.45x%2B69-2.30x%3D70.5%5C%5C%5C%5C%5CRightarrow%202.45x-2.30x%3D70.5-69%5C%5C%5C%5C%5CRightarrow%200.15x%3D1.5%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cfrac%7B1.5%7D%7B0.15%7D%5C%5C%5C%5C%5CRightarrow%20x%3D10%5C%5C%5C%5C%5Ctext%7Bplug%20this%20value%20of%20x%20in%20%28i%29%2C%20we%20get%7D%5C%5C%5C%5Cy%3D30-x%3D30-10%3D20%5C%5C%5C%5C%5Ctext%7Bso%20we%20must%20add%2010%20Pounds%20of%20trial%20mix%20that%20costs%20%5C%24%202.45%20and%2020%20pounds%20of%7D%5C%5C%5Ctext%7Bthe%20trial%20mix%20that%20costs%20%5C%242.30%7D)
That would be approximately 339638