Answer:
A option
Step-by-step explanation:
ΔMNO: ∠M=79°, ∠O=22°, then ∠N=180°-79°-22°=79°.
ΔGHK: ∠G=∠H=79°, then ∠K=180°-79°-79°=22°.
ΔPQR: ∠R=20°, ∠Q=79°, then ∠P=180°-79°-20°=81°.
ΔDEF: ∠E=82°, ∠F=22°, then ∠D=180°-82°-22°=76°.
Similar triangles have congruent angles, then
ΔMNO and ΔGHK are similar.
Answer: The answer is ![(x+7)(3x+1)=33x^2+22x+7.](https://tex.z-dn.net/?f=%28x%2B7%29%283x%2B1%29%3D33x%5E2%2B22x%2B7.)
Step-by-step explanation: Given that
![(3x2 + 22x + 7) \div(x + 7) = 3x + 1,](https://tex.z-dn.net/?f=%283x2%20%2B%2022x%20%2B%207%29%20%5Cdiv%28x%20%2B%207%29%20%3D%203x%20%2B%201%2C)
and we are to complete the following sentence:
![(x+7)(???)=???](https://tex.z-dn.net/?f=%28x%2B7%29%28%3F%3F%3F%29%3D%3F%3F%3F)
We have the following division algorithm for polynomials
![\textup{If }a(x)\times b(x)=c(x),\\\textup{then, we have }\\\\\dfrac{c(x)}{b(x)}=a(x)~~~~~\textup{or}~~~~~~c(x)\div b(x)=a(x).](https://tex.z-dn.net/?f=%5Ctextup%7BIf%20%7Da%28x%29%5Ctimes%20b%28x%29%3Dc%28x%29%2C%5C%5C%5Ctextup%7Bthen%2C%20we%20have%20%7D%5C%5C%5C%5C%5Cdfrac%7Bc%28x%29%7D%7Bb%28x%29%7D%3Da%28x%29~~~~~%5Ctextup%7Bor%7D~~~~~~c%28x%29%5Cdiv%20b%28x%29%3Da%28x%29.)
Here, a(x) = quotient, b(x) = divisor and c(x) = dividend.
Applying this rule in the given problem, we have
![\textup{since }(3x2 + 22x + 7) \div(x + 7) = 3x + 1,\\\\\textup{so, }\\\\(x+7)(3x+1)=3x^2+22x+7=0.](https://tex.z-dn.net/?f=%5Ctextup%7Bsince%20%7D%283x2%20%2B%2022x%20%2B%207%29%20%5Cdiv%28x%20%2B%207%29%20%3D%203x%20%2B%201%2C%5C%5C%5C%5C%5Ctextup%7Bso%2C%20%7D%5C%5C%5C%5C%28x%2B7%29%283x%2B1%29%3D3x%5E2%2B22x%2B7%3D0.)
Thus, the complete sentence is
![(x+7)(3x+1)=3x^2+22x+7=0.](https://tex.z-dn.net/?f=%28x%2B7%29%283x%2B1%29%3D3x%5E2%2B22x%2B7%3D0.)
An odd number of negative factors will give you a negative.
To solve this problem you need to know the law of cosines.
Three would fit cause 3 times 12 is 36 and 4 times 12 is 48 which would be too big