<span><span><span>2<span>c5</span></span>+<span>44<span>c4</span></span></span>+<span>242<span>c3</span></span></span><span><span><span>2<span>c5</span></span>+<span>44<span>c4</span></span></span>+<span>242<span>c3</span></span></span><span>=<span><span><span>2<span>c3</span></span><span>(<span>c+11</span>)</span></span><span>(<span>c+11</span><span>)</span></span></span></span>
Answer:
B
Step-by-step explanation:
no need to explain
hope it helps
Answer:
v = 6
Step-by-step explanation:
Solve for v:
-8 (8 v + 1) - 2 = -394
-8 (8 v + 1) = -64 v - 8:
-64 v - 8 - 2 = -394
Grouping like terms, -64 v - 8 - 2 = -64 v + (-8 - 2):
-64 v + (-8 - 2) = -394
-8 - 2 = -10:
-10 - 64 v = -394
Add 10 to both sides:
(10 - 10) - 64 v = 10 - 394
10 - 10 = 0:
-64 v = 10 - 394
10 - 394 = -384:
-64 v = -384
Divide both sides of -64 v = -384 by -64:
(-64 v)/(-64) = (-384)/(-64)
(-64)/(-64) = 1:
v = (-384)/(-64)
The gcd of -384 and -64 is -64, so (-384)/(-64) = (-64×6)/(-64×1) = (-64)/(-64)×6 = 6:
Answer: v = 6
Answer:

Step-by-step explanation:
![\sf{ [ (4 \frac{1}{6} + 2 \frac{1}{3} ) \div 4 \frac{1}{3}] - 1\frac{1}{2} }](https://tex.z-dn.net/?f=%20%5Csf%7B%20%5B%20%284%20%5Cfrac%7B1%7D%7B6%7D%20%20%2B%202%20%5Cfrac%7B1%7D%7B3%7D%20%29%20%20%5Cdiv%204%20%5Cfrac%7B1%7D%7B3%7D%5D%20-%20%201%5Cfrac%7B1%7D%7B2%7D%20%7D)
Convert the mixed numbers into improper fraction
![\longrightarrow{ \sf{ [ ( \frac{25}{6} + \frac{7}{3} ) \div \frac{13}{3}] - \frac{3}{2}}}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%28%20%5Cfrac%7B25%7D%7B6%7D%20%20%2B%20%20%5Cfrac%7B7%7D%7B3%7D%20%29%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%7D%20)
Add the fractions : 25 / 6 and 7 / 3
While performing addition or subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fraction.
To do so, first take the L.C.M of 6 and 3 which results to 6
![\longrightarrow\sf{ [( \frac{25 + 7 \times 2}{6} ) \div \frac{13}{3} ] - \frac{3}{2}}](https://tex.z-dn.net/?f=%20%20%5Clongrightarrow%5Csf%7B%20%5B%28%20%5Cfrac%7B25%20%2B%207%20%5Ctimes%202%7D%7B6%7D%20%29%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%20)
![\longrightarrow{ \sf{ [( \frac{25 + 14}{6} ) \div \frac{13}{3} ] - \frac{3}{2} }}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%28%20%5Cfrac%7B25%20%2B%2014%7D%7B6%7D%20%29%20%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%20%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%20%7D%7D)
![\longrightarrow{ \sf{ [ \frac{39}{6} \div \frac{13}{3}] - \frac{3}{2} }}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%5Cfrac%7B39%7D%7B6%7D%20%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%20%7D%7D)
Multiply the dividend by the reciprocal of the divisor.
Reciprocal of any number or fraction can be obtained by interchanging the position of numerator and denominator
![\longrightarrow{ \sf{ [ \frac{39}{6} \times \frac{3}{13} ] - \frac{3}{2}}}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%20%5Cfrac%7B39%7D%7B6%7D%20%20%5Ctimes%20%20%5Cfrac%7B3%7D%7B13%7D%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%7D%20)
To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term
![\longrightarrow{ \sf{ [ \frac{39 \times 3}{6 \times 13} ] - \frac{3}{2}}}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%20%5Cfrac%7B39%20%5Ctimes%203%7D%7B6%20%5Ctimes%2013%7D%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%7D%20)
![\longrightarrow{ \sf{ [ \frac{117}{78} ] - \frac{3}{2} }}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%20%5Cfrac%7B117%7D%7B78%7D%20%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%20%7D%7D)

While performing the addition or subtraction of like fractions , you just have to add or subtract the numerator respectively in which the denominator is retained same

Subtract 3 from 3

Divide 0 by 2

Hope I helped!
Best regards!