3 years ago

Its 8:05 i read for 30 mins what time did i start ?

Mathematics

2 answers:

IRISSAK [1]3 years ago

5 0

Answer:

7:35 is when you started

Step-by-step explanation:

kati45 [8]3 years ago

5 0

I agree ^ with this dude

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ASAP PLEASE HELP!! “Find the quotient”

elena-14-01-66 [18.8K]

Answer:

The quotient is $\frac{x(2x-3)}{7}$ answer (D)

Step-by-step explanation:

$\frac{2x-3}{x}$ ÷ $\frac{7}{x^{2} }$

In division with fractions we reverse the fraction after division sign and change division sign to multiplication sine

$\frac{2x-3}{x}$ × $\frac{x^{2} }{7}$

$\frac{2x-3}{1}$ × $\frac{x}{7}$

$\frac{x(2x-3)}{7}$

4 0

3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$