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3 years ago

Mr. Anderson wrote (7x9)x10 the power of 3 on the board. What is the value of that expression?

Mathematics

2 answers:

PIT_PIT [208]3 years ago

4 0

Answer:

(7x9)x10 the power of 3 is 63,000

Novay_Z [31]3 years ago

3 0

63,000000000000000000000000000000

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Prove that: (12^13–12^12+12^11)(11^9–11^8+11^7) is divisible by 3, 7, 19, and 37. The answer should be like: x^b*3*7*19*37. Also

n200080 [17]

Answer:

Step-by-step explanation:

Given (12^13–12^12+12^11)(11^9–11^8+11^7),

(12^13–12^12+12^11)(11^9–11^8+11^7) =

[(12^12)12 – 12^12 + 12^11][(11^8)11 – 11^8 + 11^7)

[(12^12)(12 – 1) + 12^11][(11^8)(11 – 1) + 11^7] =

(12^12(11) + 12^11)(11^8(10) + 11^7) =

(12^11(12x11) + 12^11)(11^7(11x10) + 11^7) =

[(12^11)(12x11 + 1)][(11^7)(11x10 + 1)] =

[(12^11)x(11^7)](12x11 + 1)(11x10 + 1) =

[(12^11)x(11^7)](133 x 111) =

[(12^11)x(11^7)](14763) =

[(12^11)x(11^7)](3x7x19x37)

From here, it is clear that the given number is divisible by 3, 7, 19 and 37.

6 0

3 years ago

In a certain Algebra 2 class of 28 students, 23 of them play basketball and 12

erica [24]

Answer:

5/14

Step-by-step explanation:

P(A or B) = P(A) + P(B) − P(A and B)

25/28 = 23/28 + 12/28 − P(A and B)

P(A and B) = 10/28

P(A and B) = 5/14

3 0

3 years ago

How do i solve this problem?

Eduardwww [97]

I think the first numerator cant be Z so I solved the question assuming it is 2;

Answer:

(x,y,z)=(1,5,7)

Hope this helps.

7 0

3 years ago

Can u guys pretty please help

erastova [34]

It should be A and D, no promises

7 0

3 years ago

BRAINLIEST for the correct answer!

Vladimir [108]

Answer:

answer is A

Step-by-step explanation:

(3 $x^{2}$ +5x+1) ( $x^{2}$ −2x+4)

=(3 $x^{2}$ +5x+1)( $x^{2}$ +−2x+4)

=(3 $x^{2}$ )( $x^{2}$ )+(3 $x^{2}$ )(−2x)+(3 $x^{2}$ )(4)+(5x)( $x^{2}$ )+(5x)(−2x)+(5x)(4)+(1)( $x^{2}$ )+(1)(−2x)+(1)(4)

=3 $x^{4}$ −6 $x^{3}$ +12 $x^{2}$ +5 $x^{3}$ −10 $x^{2}$ +20x+ $x^{2}$ −2x+4

=3 $x^{4}$ − $x^{3}$ + 3 $x^{2}$ + 18x +4

4 0

3 years ago