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masya89 [10]
3 years ago
12

Why is the distributive property appropriate for two-digit multiplication?

Mathematics
1 answer:
Rina8888 [55]3 years ago
4 0
Here's an example:

25\times93=(20+5)\times(90+3)
25\times93=20\times(90+3)+5\times(90+3)
25\times93=20\times90+20\times3+5\times90+5\times3
25\times93=1800+60+450+15
25\times93=2325
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PLEASE HELP ME!! (50 COINS) (i rlly need help so im giving coins away PLEASE ANSWER IMMEDIATELY!!!)
Archy [21]

Hello from MrBillDoesMath!

Answer:

4( x + 1.5)^2 + 0  

Discussion:

4x^2 + 12x + 9 =                => factor "4" from first 2 terms

4 (x^2 + 3x) + 9 =               => complete the square, add\subtract (1.5)^2

4(x^2 + 3x + (1.5)^2)  - 4 (1.5)^2 + 9  =

4 ( x + 1.5)^2 + ( 9 - 4(1.5)^2) =           => as (1.5)^2 = 2.25

4 ( x + 1.5)^2 + ( 9 - 4(2.25)) =        => as 4 ( 2.25) = 9

4 ( x+ 1.5)^2 + 0


Thank you,

MrB

5 0
2 years ago
9 - 10x = 2x + 1 - 8x
Furkat [3]

Answer:

x=2

Step-by-step explanation:

9 - 10x = 2x + 1 - 8x

9-10x=-6x+1

Subtract 9 from both sides:

9-10x-9=-6x+1-9

-10x=-6x-8

Add 6x to both sides:

-10x+6x=-6x-8+6x

-4x=-8

Divide both sides by -4:

\frac{-4x}{-4}=\frac{-8}{-4}

x=2

8 0
3 years ago
Read 2 more answers
Write the expression without the fraction bar.
kirill115 [55]
For problems like this you can move the y's on the bottem up top by fliping the sign of the exponent (in this case 7 to -7) and MULTIPLYING it with EVERYTHING on top, then because of properties of multplication you can add the exponents to combine them into one term (in this case add 3 and -7 to get -4) \frac{ y^{3} }{y^{7}}=y^3 *y^{-7} = y^{3+-7}=y^{3-7}=y^-4 If any of this does not make sense let me know and ill try to clarify better
5 0
2 years ago
Read 2 more answers
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
How many days will it take Yertle the Turtle to climb out of the bottom of a 15-feet deep well if he climbs up 2 1/2 feet each d
sergiy2304 [10]

Answer:

It will take him 7 ½ days.

Step-by-step explanation:

He goes up 2 ½ but falls back down ½ so you take the two and do 15/2 and get 7.5 which is 7 ½ days.

4 0
3 years ago
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