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3 years ago

Word form 9,000,900,009

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

7 0

Nine billion, nine hundred thousand, nine.

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Is algebra.

Dmitry [639]

Factor By Grouping
GCF

7 0

3 years ago

Find the number of five-digit numbers which are not divisible by 5?

storchak [24]

Answer:

72000

Step-by-step explanation:

First five digit number is 10000

And the last five digit number is 99999.

There are 99999-10000+1 five digits in all.

That simplifies to 89999+1=90000.

Any number whose units digit is 5 or 0 is divisble by 5.

10000 is the first five digit number that is divisble by 5.

10000+5k where k is a positive integer can be used to find any after.

The last five digit number divisible by 5 is 99995. What is k then?

10000+5k=99995

5k=89995

k=89995/5

k=17999

How many 5 digit numbers are divisble by 5?

So the first one happens at k=0 and the last one happens at k=17999.

How many numbers is that? 17999+1=18000.

So the number of 5 digit numbers that aren't divisble by 5 is 90000-18000=72000

7 0

3 years ago

Which fractions are equivalent to 3/4? check all tha apply.

Alik [6]

Answer:

6/8, 9/12, and 15/20

Step-by-step explanation:

just do 3 divided by 4 (3/4) and it equals 0.75 then do each and everyone like 6 divided by 8 (6/8) so it equals 0.75

7 0

3 years ago

Find the gcf of 15a and 28b^2. 9ab^2 0 3 1

lions [1.4K]

Answer:

GCF(15a; 28b²) = 1

5 0

4 years ago

Solve the system by elimination. − 2 x + 2 y + 3 z = 0 − 2 x − y + z = − 3 2 x + 3 y + 3 z = 5

taurus [48]

Answer:

$x=1, y=1, z=0$

Step-by-step explanation:

we have

$-2x+2y+3z=0$ ----> equation A

$-2x-y+z=-3$ -----> equation B

$2x+3y+3z=5$ ----> equation C

adds equation A and equation C

$-2x+2y+3z=0\\2x+3y+3z=5\\-------\\2y+3y+3z+3z=0+5$

$5y+6z=5$ -----> equation D

adds equation B and equation C

$-2x-y+z=-3\\2x+3y+3z=5\\--------\\-y+3y+z+3z=-3+5$

$2y+4z=2$ -----> equation E

we have the system

$5y+6z=5$ -----> equation D

$2y+4z=2$ -----> equation E

Multiply equation E by -1.5 both sides

$-1.5(2y+4z)=-1.5(2)$

$-3y-6z=-3$ -----> equation F

adds equation D and equation F

$5y+6z=5\\-3y-6z=-3\\---------\\5y-3y=5-3\\2y=2\\y=1$

<em>Find the value of z</em>

substitute the value of y in equation E (or substitute in equation D)

$2(1)+4z=2$

$2+4z=2$

$4z=0$

$z=0$

<em>Find the value of x</em>

substitute the value of y and z in equation A (or equation B or C)

$-2x+2(1)+3(0)=0$

$-2x+2=0$

$-2x=-2$

$x=1$

therefore

The solution of the system of equations is

$x=1, y=1, z=0$

4 0

4 years ago