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STALIN [3.7K]
3 years ago
12

OMG IM SO CONFUSED HELPPPP

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
8 0

Answer:

5.4 is rational. 5.4 divide by 2 is 2.45 5.3 and 5 3 are irrational, as it cant be divided easily

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Solve the system of equations by graphing.
valentinak56 [21]
The answer is A.

The way you know this is by graphing the first function.

The second function is in a different format, so you just have to isolate Y. 

-1/3x + y = -1
y= 1/3x -1  OR y= -1 + 1/3x

both of the functions have the same slope, but different y-intercepts. therefore, they'll never touch. making them infinite solutions.

it's also A because both of the slopes are positive, the slopes in graph D are negative.

So, the answer is A.

Hope this helps! xx


4 0
3 years ago
There are 4 red cars and 8 black cards in a bag. How is it likely that you will randomly draw a blue card
Verdich [7]

Answer:

C) Impossible

Step-by-step explanation:

Since there are no blue cards, the probability that you'd draw one is 0/12, which means that there is no chance you'll draw one.

4 0
3 years ago
Which value of x makes the equation 1.25(6x - 20) = 50 true?
bixtya [17]
D. X = 10

6x-20= 40

6x=40+20

6x=60

X=10
4 0
2 years ago
Selene's checking account balance was -$33, and then she withdrew $42 more. What is her checking account balance after the withd
Rainbow [258]

Answer:

-75

Step-by-step explanation:

WIthdrew = Negative

-33 - 42 = -75

7 0
2 years ago
Read 2 more answers
What is the smallest multiple of 18 of the form 2A945B, where A and B are digits?
musickatia [10]

Answer:

219456

Step-by-step explanation:

It must end in 0,2,4,6,8 (divisible by 2 rule) and the digits must sum to a multiple of 9(divisible by 9 rule).

The digit sum is 2+A+9+4+5+B=20+A+B.

We want to make A as low as possible because it's a higher digit than B.

If A is as low as possible, we have that the sum is 20+0+7=27. Uh-oh. B has to be even! So A=0 doesn't work.

If A=1, we have that the sum is 20+1+6=27. Yay!

So, the answer is \boxed{219456}

6 0
2 years ago
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