All categories

2 years ago

Bonita said that the product of 5/6 times 1 2/3 is 7/3 how can you tell her answer is wrong?

Mathematics

2 answers:

kotegsom [21]2 years ago

6 0

The answer is really 1 5/6 because 5/6 + 2/3=5/6 NOT 7/3 so of course she is wrong

iren [92.7K]2 years ago

4 0

Answer:

25/ 18 is not equal to 7/3

Step-by-step explanation:

Bonita said that the product of 5/6 times 1 2/3 is 7/3

( 5/6)x ( 1 ⅔) = 7/3

( 5/6) x( 5/3) = 7/3

5/6 x 5/3 = 7/3

Thus, her answer is wrong 25/ 18 is not equal to 7/3

You might be interested in

Sansa wrote this expression:

NISA [10]

Answer:

34 is the answer

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Log 300 =<br> log 3.26 =<br> log 10,000 =<br> log 20 =

ehidna [41]

Answer:

Log300=2.4771

Log3.26=0.5132

Log10000=4

Log20=1.3010

Step-by-step explanation:

log300=2.4771

Log3.26=0.5132

Log10000=4

Log20=1.3010

4 0

3 years ago

Why are (1,2) and(2,1) not equal ordered pair?

vladimir1956 [14]

Answer:

The pairs ( 1 , 2 ) and ( 2 , 1 ) are not equal because their respective elements are not equal.

Step-by-step explanation:

The pair of elements which are in specific order is called an ordered pair. The pair ( 1 , 2 ) is not same as the pair ( 2 , 1 ). In the pair ( 1 , 2 ) 1 is in the first position and 2 is in the second position. In the pair ( 2 , 1 ), 2 is in the first position and 1 is in the second position.

Two ordered pairs ( a , b ) and (c , d ) are said to be equal if a = c and b = d. We write ( a , b ) = ( c , d ).

Hope I helped!

Best regards!!

8 0

2 years ago

Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract

anyanavicka [17]

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$ and that equality is true as long as $|r| < 1$ .

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$ and $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$