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Digiron [165]
3 years ago
7

What is one third plus one seventh

Mathematics
1 answer:
Kryger [21]3 years ago
8 0
To do this, change into common denominators.  The common denominator is 21 so change the problem from 1/3 + 1/7 to 7/21 + 3/21.  This gives you 10/21.  It is already in simplest form.
You might be interested in
Simplify each expression. Assume that all variables are positive.
kozerog [31]
Q1. The answer is  \frac{8x^{3}y^{6}  }{27}

( \frac{16 x^{5} y^{10}}{81x y^{2} } )^{ \frac{3}{4} }= ( \frac{16}{81}* \frac{ x^{5} }{x}* \frac{ y^{10} }{y^{2}}   )^{ \frac{3}{4} } \\  \\ 
  \frac{ x^{a} }{ x^{b} }= x^{a-b}  \\  \\ 
( \frac{16}{81}* \frac{ x^{5} }{x}*\frac{ y^{10} }{y^{2}}   )^{ \frac{3}{4} }}=( \frac{16}{81 }* x^{5-1}* y^{10-2})^{ \frac{3}{4} }=( \frac{16}{81 }* x^{4}* y^{8})^{ \frac{3}{4} }= \\  \\ = (\frac{16}{18} )^{ \frac{3}{4} }*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=
\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\  \\ 
 (x^{a} )^{b} = x^{a*b}  \\  \\ 
\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } =  \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} = 
= \frac{8x^{3}y^{6}  }{27}

Q2. The answer is 1/16

(-64) ^ \frac{-2}{3} =(-1* 2^{6} ) ^ \frac{-2}{3}=(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} \\\\x^{-a} =  \frac{1}{ x^{a} } \\\\(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} = \frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{(2^{6})^ \frac{2}{3}} \\  \\  (x^{a} )^{b}=x^{a*b} \\\\x^{ \frac{a}{b} = \sqrt[b]{ x^{a} } }  \\  \\ 

\frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{2^{6*\frac{2}{3}}} = \frac{1}{ \sqrt[3]{(-1)^{2} } } * \frac{1}{ 2^{4} } =  \frac{1}{ \sqrt[3]{1} } * \frac{1}{16} = \frac{1}{1} * \frac{1}{16}= \frac{1}{16}


Q3. The answer is a^{ \frac{7}{6} }

a^{ \frac{2}{3} } * a^{ \frac{1}{2} }  \\  \\ 
 x^{a}* x^{b}  =x^{a+b}  \\  \\ 
a^{ \frac{2}{3} } * a^{ \frac{1}{2} }= a^{ \frac{2}{3} + \frac{1}{2} } =a^{ \frac{2*2}{3*2} + \frac{1*3}{2*3} }=a^{ \frac{4}{6} + \frac{3}{6} }=a^{ \frac{4+3}{6} }=a^{ \frac{7}{6} }
7 0
3 years ago
Ten more than x times 6 is 11 more than y times 4
Elza [17]
I’m supposing you are writing an equation for this. The equation could be written as 6x+10=4y+11.
3 0
3 years ago
Read 2 more answers
Plz helpppppppp me thxs​
Alenkasestr [34]

Answer:thxxxxxxs for the points

Step-by-step explanation:

8 0
2 years ago
Solve the equation 2^(2x+1)+8=17×2^x.
myrzilka [38]
Oooh, looks fun

ok
erm
we might want to know some properties
(x^m)(x^n)=x^{m+n}
(a^b)^c=a^{bc}
if a^m=a^n where a=a, then assume m=n

2^{2x+1}=(2^{2x})(2^1)
so
(2^{2x})(2)+8=17(2^x)
(2^{2x})(2)+2^3=17(2^x)
(2^x)^2(2)+2^3=17(2^x)
minus 17*2^x both sides
(2^x)^2(2)+2^3-17(2^x)=0
use u subsitution, u=2ˣ
(u)^2(2)+2^3-17(u)=0
solve 2u^2+8-17u=0
or
2u^2-17u+8=0
ac method
2 times 8=16
what 2 number multiply to get -17 and add to get 16
-16 and -1
2u²-1u-16u+8=0
(2u²-1u)+(-16u+8)=0
u(2u-1)+(-8)(2u-1)=0
(u-8)(2u-1)+0
u-8=0
u=8

2u-1=0
2u=1
u=1/2

now
u=2ˣ

8=2ˣ
2^3=2^x
3=x

and
\frac{1}{2}=2^x
2^{-1}=2^x
-1=x

x=-1 and 3
neat problem
8 0
3 years ago
A doctor's office schedules 15 minutes for each patient and a half hour for the doctor's lunch at 11:30 a.m. If the doctor start
Ierofanga [76]

9514 1404 393

Answer:

  24

Step-by-step explanation:

If we reschedule the doctor's half-hour break so that s/he quits work at 3:00 pm, s/he will have worked 6 straight hours from 9 am to 3 pm. Seeing 4 patients each hour, the doctor will have seen ...

  (6 h) × (4 pt/h) = 24 pt

The doctor saw 24 patients by 3:30 pm.

6 0
3 years ago
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