All categories

2 years ago

What is the product of 5.9 x 103 and 5.9 x 104 expressed in scientific notation?

Mathematics

1 answer:

stiv31 [10]2 years ago

3 0

The answer to this question would be 5687

You might be interested in

Student scientific calculator 3×10^5 is displayed as 3 EE 5 in that case how was the panic of 80,000 and 5000 be displayed on Th

faltersainse [42]

Hello,
We know that
3×10^5= 3 EE 5
Change 80,000 to the scientific notation
8×10^4
Doing the same like 3×10^5, I can get 8 EE 4
Change 5000 to the scientific notation
5×10^3
Then it will appeared as 5 EE 3. Hope it help!

4 0

3 years ago

14 : ( 8 - 1 ) + 6 + 7 : 7 - ( 12 - 10 ) : 2 =

goldfiish [28.3K]

The correct answer is 14.

7 0

3 years ago

What is x in this Math problem 2/3x=27

jek_recluse [69]

Answer: x = 40.5

Step-by-step explanation:

Simply divide 27/(2/3) to get 40.5

Hope it helps <3

4 0

3 years ago

Read 2 more answers

In 1998, Cathy's age is equal to the sum of the four digits in the year of her birthday, then how old was Cathy in 1998?

ycow [4]

Answer:

Cathy was born in 1980 and she was 18 years old in 1998

Step-by-step explanation:

Equations

This is a special type of equations where all the unknowns must be integers and limited to a range [0,9] because they are the digits of a number.

Let's say Cathy was born in the year x formed by the ordered digits abcd. A number expressed by its digits can be calculated as

$x=1000a+100b+10c+d$

In 1998, Cathy's age was

$1998-(1000a+100b+10c+d)$

And it must be equal to the sum of the four digits

$1998-(1000a+100b+10c+d)=a+b+c+d$

Rearranging

$1998=1001a+101b+11c+2d$

We are sure a=1, b=9 because Cathy's age is limited to having been born in the same century and millennium. Thus

$1998=1001+909+11c+2d$

Operating

$88=11c+2d$

If now we try some values for c we notice there is only one possible valid combination, since c and d must be integers in the range [0,9]

c=8, d=0

Thus, Cathy was born in 1980 and she was 18 years old in 1998. Note that 1+9+8+0=18

5 0

3 years ago

Use the following steps to prove that log b(xy)- log bx+ log by.

borishaifa [10]

Answer with Step-by-step explanation:

a. $x=b^p$

$y=b^q$

Taking both sides log

$log x=plog b$

Using identity: $logx^y=ylogx$

$p=\frac{logx}{log b}=log_b x$

Using identity: $log_x y=\frac{log y}{log x}$

$log y=qlog b$

$q=\frac{log y}{log b}=log_b y$

b. $xy=b^pb^q$

We know that

$x^a\cdot x^b=x^{a+b}$

Using identity

$xy=b^{p+q}$

c. $log_b(xy)=log_b(b^{p+q})$

$log_b(xy)=(p+q)log_b b$

Substitute the values then we get

$log_b(xy)=(log_b x+log_b y)$

By using $log_b b=1$

Hence, $log_b(xy)=log_b x+log_b y$

3 0

3 years ago