All categories

3 years ago

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:

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