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Scrat [10]
3 years ago
7

Simplify the expression

0%7B%7D%5E%7B2%7D%20%20-%206y%20%3D%20%20" id="TexFormula1" title="3y { }^{2} + 9y + 12 + 4y {}^{2} - 6y = " alt="3y { }^{2} + 9y + 12 + 4y {}^{2} - 6y = " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Ber [7]3 years ago
7 0
The simplified equation = 7y^2+3y+12
laila [671]3 years ago
3 0
7y2+3y+12= I don’t know because that’s all you put
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Please help please please help please
zalisa [80]
Spinning on Y out of three letters (G, I and Y) = 1/3
then either an easy or hard = 1/2

so
The probability that the spinner will stop on the letter Y and then easy
= 1/3 x 1/2
= 1/6
= 0.17

answer:
0.17

5 0
3 years ago
Need help on 9 don’t know if I did it right or not but I need help :p
solmaris [256]

Answer:

9b) -8 = y

9a) 72° = <em>m</em>∠<em>ABC</em>

Step-by-step explanation:

Since you have an angle trisector, in this case, <em>m</em>∠<em>CBE</em><em> </em>≅ <em>m</em>∠<em>DBA,</em><em> </em>therefore you <em>set</em><em> </em><em>x</em><em> </em>equal to 8, plus, according to Morley's Trisector Theorem, all three angles form an equilateral triangle, so <em>m</em>∠<em>BOC</em><em> </em>also has to equal 24°:

{8}^{2}  - 5(8) = 64 - 40 = 24 \\  \\ 6(8) - 24 = 48 - 24 = 24 \\  \\ 2(8) - y = 24 >> 16 - y = 24 >>  -8 = y

Then, <em>m</em>∠<em>ABC</em><em> </em>comes from multiplying 3 by 24 [three <em>twenty-</em><em>four</em>'s], which results in 72°.

I am joyous to assist you anytime.

4 0
3 years ago
HELPPPPPPPPPPPPPPPPPP
Nookie1986 [14]
162 - 9 = 153. 
d = 153.
Diego is old.
3 0
3 years ago
Read 2 more answers
A bicycle lock has a four-digit code. The possible digits, 0 What is the probability that the lock code will begin with
nika2105 [10]

Answer:

0.1-0.6

Step-by-step explanation:

First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.

The first digit is any of {0, 1, 2...,9}, that is 10 possibilities

the second digit is any of the remaining 9, after having picked one. 

and so on...

so in total there are 10*9*8*7 = 5040 codes.

a. What is the probability that the lock code will begin with 5?

Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.

Thus, there are 1*9*8*7=504 codes which start with 5.

so 

P(first digit is five)=

b. What is the probability that the lock code will not contain the number 0? 

from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}

from which we can form in total 9*8*7*6 codes which do not contain 0.

P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6

Answer:

0.1 ; 0.6

8 0
3 years ago
Read 2 more answers
Jeremiah has to spend $40 to buy candy. He then sold candy for $1.50 per piece. If Jeremiah was able to sell 60 pieces of the ca
Nadya [2.5K]

Answer:

He made $50

Step-by-step explanation:

1.5 times 60 equals 90, but seeing how he spent $40 to buy the candy he only ends up with $50 in profits

7 0
3 years ago
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