4# is a two digit number ,where # represents the digit at ones place . 7999 ÷ 4# lies between ---------

Mathematics

1 answer:

Hitman42 [59]3 years ago

4 0

Answer:

$7999 \div 4\#$ lies between 163.245 and 199.975

Step-by-step explanation:

Given

2 digit = 4#

Required

The range of $7999 \div 4\#$

Let

$\# = 0$ --- the smallest possible value of #

So:

$7999 \div 40 = 199.975$

Let

$\# = 9$ --- the largest possible value of #

So:

$7999 \div 49 = 163.245$

<em>Hence, </em> $7999 \div 4\#$ <em> lies between 163.245 and 199.975</em>

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Find the quotient.

posledela

Answer :

$B. 16 {a}^{2}c$

step-by-step explanation :

$48 {a}^{3}b{c}^{2} \div 3abc$

This can be rewritten as:

$\frac{ 48 {a}^{3}b {c}^{2} }{3abc}$

Now,

$\frac{48}{3}=16$

The law of indices states that:

$\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

It implies that, when dividing two expressions with the same bases, repeat one of the bases and subtract the exponents.

Therefore,

$\frac{ {a}^{3} }{a}= {a}^{3 - 1} = {a}^{2}$

$\frac{b}{b} = {b}^{1 - 1} = {b}^{0} = 1$

Note: Any non-zero number exponent zero is 1

Also

$\frac{ {c}^{2} }{c} = {c}^{2 - 1} = {c}^{1} = c$

Hence:

$\frac{ 48 {a}^{3}b {c}^{2} }{3abc} = 16 {a}^{2}c$