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Anna71 [15]
3 years ago
8

The vertex form of the equation of a parabola is y = 3(x-4)2 +13. What is the

Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
5 0
It is d y-1x2-Bc+29 I say that bc it says that in the answer up above
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5/12 0.12 19/50 least to greatest
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0.12, 19/50, 5/12. Convert the fractions into decimals by dividing the numerator by the denominator to find the answer to questions like this more easily.
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3 years ago
A coordinate grid with 2 lines. The first line is labeled y equals negative StartFraction 7 over 4 EndFraction x plus StartFract
LekaFEV [45]

Answer:

Question 1. (2.2, -1.4)

Question 2. (1.33, 1)

Step-by-step explanation:

Equations for the given lines are

-----(1)

It is given that this line passes through two points (0, 2.5) and (2.2, 1.4).

------(2)

This equation passes through (0, -3) and (2.2, -1.4).

Now we have to find a common point through which these lines pass or solution of these equations.

From equations (1) and (2),

x =

x = 2.2

From equation (2),

y = -1.4

Therefore, solution of these equations is (2.2, -1.4).

Question 2.

The given equations are y = 1.5x - 1 and y = 1

From these equations,

1 = 1.5x - 1

1.5x = 2

x =

Therefore, the solution of the system of linear equations is (1.33, 1).

5 0
3 years ago
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If m∠2 = 137 and m∠P = 22, what is m∠O?
Katyanochek1 [597]

Answer:

The answer is 21

Step-by-step explanation:

This should help you

8 0
1 year ago
You owe $1,534.09 on a credit card that has an interest rate of 15.7% APR. You decide to put $250.00 into a savings account earn
Brilliant_brown [7]

you shouldnt answer if you really dont know you noodle head , the answer is 2.54

6 0
3 years ago
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 
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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
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