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kolbaska11 [484]
3 years ago
13

Solve using the quadratic equation. Round your solution to the nearest hundredth of 2x^2-3x+4

Mathematics
2 answers:
ch4aika [34]3 years ago
8 0

Answer:

2x2+3x−4=0 is a quadratic equation in the form ax2+bx+c, where a = 2, b = 3, and c = − 4.

Step-by-step explanation:

Hope this helps

ANTONII [103]3 years ago
5 0

Answer:

brain list me please

Step-by-step explanation:

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6+5 x 2 +5+ = <br><br> 67 +84 - 12 x 4 = 16
lozanna [386]

Answer:

6+(5 x 2)+5=21

the way you wrote this was kinda weird so I don't know what you expected as an answer

Step-by-step explanation:

5 x 2=10

10+6=16

16+5=21

21

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3 years ago
1+1 abcdefghijklmnopqrstuvwxyandz
skad [1K]

Answer:2


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3 years ago
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Answer:

mean= 3.28

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3 years ago
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What is a three digit number thats an odd multiple of three and the product of its digits is 24 and is larger than 15 squared?
Reika [66]
The only way 3 digits can have product 24 is 
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4 
To be divisible by 3 the sum of the digits must be divisible by 3.
1+  3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3. 

So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.

Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.

The others must end in 3. 

They must be greater than 152 which is 225. So the

First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.

The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.  
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
7 0
3 years ago
What are the coordinates of the terminal point determined by T=20pi/3
Serga [27]

Answer:

t=5x=3049

Step-by-step explanation:

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2 years ago
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