We are given that
.
Substitute x with 5-x, then the above equation becomes:
, that is
So, we have the following system of equations:
i)
ii)
multiply the first equation by -4, so that we eliminate f(5-x)'s
i)
ii)
adding the 2 equations side by side we have:
expanding the binomial, and collecting same terms we have:
dividing by -5:
dividing by 3:
Answer:
Use PEMDAS. Parenthesis. Exponents. Multiplication. Division. Addition. Subtraction. there are no parenthesis so that doesn't matter. there are no exponents. but there is multiplication. 8×7 = 56. so now the equation is 56+4×2-11-7×4. you still have multiplication so now we do 4×2 which is 8. the equation is now 56+8-11-7×4. you still have multiplication so now do 7×4 which is 28. the equation is now 56+8-11-28. there is no multiplication. or division. next is addition. 56+8 which is 64. now the equation is 64-11-28. next you do subtraction. 64-11 which is 53. now the equation is 53-28 which is 25. 25 is the answer. I hope this helped ☺️
Answer:
The mother (Rhoda) is 46 years old.
The daughter (Tenica) is 18 years old
Step-by-step explanation:
Let the age of the mother (Rhoda) be m
Let the age of the daughter (Tenica) be d.
The sum of Rhonda and her daughter Tenica’s age is 64. This can be written as:
m + d = 64 ... (1)
The difference in their ages is 28. This can be written as:
m – d = 28 ... (2)
From the above illustrations, the equation obtained are:
m + d = 64 ... (1)
m – d = 28 ... (2)
Solving by elimination method:
Add equation 1 and 2 together
. m + d = 64
+ m – d = 28
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
2m = 92
Divide both side by 2
m = 92/2
m = 46
Substitute the value of m into any of the equation to obtain the value of d. Here, we shall use equation 1
m + d = 64
m = 46
46 + d = 64
Collect like terms
d = 64 – 46
d = 18
Therefore, the mother (Rhoda) is 46 years old and the daughter (Tenica) is 18 years old.
is simply the difference of both amounts, but firstly let's convert the mixed fractions to improper, and subtract.
![\bf \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\\\ \stackrel{mixed}{6\frac{7}{16}}\implies \cfrac{6\cdot 16+7}{16}\implies \stackrel{improper}{\cfrac{103}{16}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{Jessie}{\cfrac{103}{16}}-\stackrel{Bryce}{\cfrac{9}{2}}\implies \stackrel{\textit{our LCD is 16}}{\cfrac{(1)103-(8)9}{16}}\implies \cfrac{103-72}{16}\implies \cfrac{31}{16}\implies 1\frac{15}{16}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B7%7D%7B16%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%2016%2B7%7D%7B16%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B103%7D%7B16%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cstackrel%7BJessie%7D%7B%5Ccfrac%7B103%7D%7B16%7D%7D-%5Cstackrel%7BBryce%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bour%20LCD%20is%2016%7D%7D%7B%5Ccfrac%7B%281%29103-%288%299%7D%7B16%7D%7D%5Cimplies%20%5Ccfrac%7B103-72%7D%7B16%7D%5Cimplies%20%5Ccfrac%7B31%7D%7B16%7D%5Cimplies%201%5Cfrac%7B15%7D%7B16%7D)
