All categories

3 years ago

1000000+200000=? fun for me

Mathematics

1 answer:

Arturiano [62]3 years ago

6 0

Answer:

1,200,000

Step-by-step explanation:

hope i helped

You might be interested in

Plz help , need help on 6, 7 &8

tatyana61 [14]

I'm pretty sure number 6 is 9.23 in

6 0

4 years ago

Read 2 more answers

Please help anyone

Iteru [2.4K]

Answer:

p = 30

Step-by-step explanation:

EFGH is a cyclic quadrilateral ( all vertices lie on a circle )

The opposite angles of a cyclic quadrilateral are supplementary, thus

p + 5p = 180

6p = 180 ( divide both sides by 6 )

p = 30

4 0

3 years ago

What is the slope of the line through (-9,6)(−9,6)

Vika [28.1K]

Answer:

Well I think It's undefined

Step-by-step explanation:

m = y2 - y1/x2 - x1

= 6-6/-9-(-9)

= 6-6/-9+9

m = 0/0

5 0

3 years ago

Read 2 more answers

67. The Goldbach Conjecture. Recall that a prime number is a

Korvikt [17]

Answer:

<u>Test the conjecture for 10 even numbers</u>

10 = 3 + 7

12 = 5 + 7

14 = 7 + 7

16 = 5 + 11

18 = 5 + 13

20 = 7 + 13

22 = 11 + 11

24 = 11 + 13

26 = 13 + 13

28 = 11 + 17

<u>Do you think the conjecture is true? Why or why not?</u>

I believe the conjecture is true because the sum of two odd numbers is an even number and you can find two prime odd numbers for every even number whose sum is that even number.

4 0

3 years ago

Which expressions are equivalent to \dfrac{4^{-3}}{4^{-1}} 4 −1 4 −3 start fraction, 4, start superscript, minus, 3, end super

Slav-nsk [51]

Answer:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}$

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}$

Step-by-step explanation:

Given

$\dfrac{4^{-3}}{4^{-1}}$

Required

Choose equivalent expressions

Choosing the first answer:

$\dfrac{4^{-3}}{4^{-1}}$

Split expressions

$4^{-3} * \frac{1}{4^{-1}}$

Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

$\frac{1}{4^3} * \frac{1}{4^{-1}}$

Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

$\frac{1}{4^3} * \frac{1}{1/4}$

$\frac{1}{4^3} * \frac{4^1}{1}$

$\frac{4^1}{4^3}$

Hence:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}$

Choosing the second:

$\dfrac{4^{-3}}{4^{-1}}$

Apply law of indices: $(\frac{a^m}{a^n} = a^{m-n})$

So,

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3-(-1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3+1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-2}$

Apply law of indices: $(a^{-b} = \frac{1}{a^b})$

So:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}$

4 0

3 years ago