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Basile [38]
3 years ago
10

Look for Patterns Find the missing numbers in the following pattern. 1,3,9, .81 .​

Mathematics
2 answers:
ehidna [41]3 years ago
5 0

Answer:

27 after 81 is 243

Step-by-step explanation:

The pattern is that you multiply 3 each time.

jasenka [17]3 years ago
4 0

Answer:

the pattern id going 1*3=3 3*3=9 9*9=81

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URGENT !!!!!!!!!!!! Please answer correctly !!!!!!!!! Will<br> Mark Brianliest !!!!!!!!!!!!!!!!
mina [271]

Answer:

MLH is congruent to JKI

LNK is congruent to HNI

I think that's how you're supposed to answer it...

3 0
3 years ago
Scott found a rare comic book on an online auction site priced at 35 Australian dollars. If the exchange rate is $1 U.S=$1.09 Au
adell [148]
1 US = 1.09 Australian

divide by 1.09 on both sides to find how much a 1 Australian dollar is worth.

1/1.09 = 1 Australian dollar
1 Australian dollar = 0.92 US dollar

mulitply 35 on both sides of the equation

35 Australian = 35 (0.92) = 32.2 US dollars
8 0
3 years ago
How to solve y=-x+4; y=2x-8 using substitution?
Novosadov [1.4K]
In this problem why is the easiest to substitute.
Just set the equations equal to each other because they both equal y

-x + 4 = 2x - 8
(You want to get all the x's to one side so add x to each side)
4 = 3x - 8
(next add 8 to both sides)
12 = 3x
(lastly divide 3 from both sides)
4 = x
You have got your x so now it is time to find that y just replace x to either of the two equations
y = 2(4) -8   or  y = -(4) + 4
y = 8 - 8          y = 0
y = 0
Final Answer:  y=0, x=4
4 0
4 years ago
On Monday, the ABC Company stock was at $12.80 per share. On Tuesday it gained $3.42, and then lost $9.70 on Wednesday. What is
solong [7]

Answer:

Step-by-step explanation:

12.80+3.42= 16.22

16.22-9.70= 6.52 that's your answer

6 0
3 years ago
How can I factorise (2x cubed - 5x + 3) ?
PilotLPTM [1.2K]
2x^3-5x+3=2x^3-2x-3x+3\\\\=2x(x^2-1)-3(x-1)\\\\=2x\underbrace{(x^2-1^2)}_{use\ (*)}-3(x-1)\ \ \ |(*)\ a^2-b^2=(a-b)(a+b)\\\\=2x\underbrace{(x-1)}_{(**)}(x+1)-3\underbrace{(x-1)}_{(**)}\\\\=(x-1)[2x(x+1)-3]\\\\=(x-1)\underbrace{(2x^2+2x-3)}_{(***)}\\\\(***)\ 2x^2+2x-3\to a=2;\ b=2;\ c=-3\\\\x=\frac{b^2\pm\sqrt{b^2-4ac}}{2a}

therefore\\x=\frac{-2\pm\sqrt{2^2-4(2)(-3)}}{2(2)}=\frac{-2\pm\sqrt{4+24}}{4}=\frac{-2\pm\sqrt{28}}{4}=\frac{-2\pm\sqrt{4\cdot7}}{4}=\frac{-2\pm2\sqrt7}{4}\\\\=\frac{-1\pm\sqrt7}{2}\\\\so,\ the\ answer:\\\\(x-1)\cdot2\left(x-\frac{-1-\sqrt7}{2}\right)\left(x-\frac{-1+\sqrt7}{2}\right)\\\\=\boxed{(x-1)(2x+1+\sqrt7)\left(x+\frac{1-\sqrt7}{2}\right)}=\boxed{\frac{1}{2}(x-1)(2x+1+\sqrt7)(2x+1-\sqrt7)}
7 0
3 years ago
Read 2 more answers
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