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Nataly [62]
3 years ago
12

PLEASE PLEASE help with math homework-its due soon and I don't get it. thanks so much!

Mathematics
2 answers:
ollegr [7]3 years ago
7 0
I’m pretty sure you have to substitute

6.75-6(1)= .75

6.75-6(2)= -5.25

6.75- 6(5)= -23.25

6.75- 6(-3.25)= 26.25

4/4 +3(4)= 13

8/4 +3(8)= 26

-12/4 +3(-12)= -39

-1/4+3(-1)= -13/4

Hope this helps :)
Roman55 [17]3 years ago
5 0
You just change the number with the word
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lapo4ka [179]
This is a rectangle... Did you draw the points?
 
Find
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- and the length of the segment from (-20,-15) to (-15,-20).

These are the length and width of the rectangle; multiply both quantities.

The width is found with the formula:
\sqrt{x_1 - x_2)^2 + (y_1 - y_2)^2} =  \sqrt{((-15) - (-20))^2 + ((-20) - (-15))^2}
The length is \sqrt{x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{((-15)-(0))^2+((-20)-(-5))^2} 

4 0
3 years ago
Jason's long jump was 98 inches. Dougs long jump was3 yards. How much longer was Dougs jump than Jason's jump?
kari74 [83]
Convert 3 yards to inches.

1 yard = 36 inches

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3 0
3 years ago
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Round 7095075 to the nearest million
nignag [31]
<h2>Answer: 7,000,000</h2>

Step-by-step explanation:

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4 0
3 years ago
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3 years ago
Over which interval(s) is g(x) = x2 + x - 12 positive?
VikaD [51]

Answer:

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Step-by-step explanation:

Factor and set = to 0

x^{2}  + x - 12\\( x + 4)(x - 3) = 0

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The two numbers would divide a number line into 3 intervals.  Pick a value in one of the intervals and put it in the original expression.  If it makes the function positive, then all the values in that interval make the function positive.  If the value you picked makes the function negative, then the values in the other intervals will make the function negative.  Let's pick the value of 0 and substitute it into the function

We get 0^{2} + 0 - 12 = -12 which is not positive.  Therefore, all the values between -4 and 3 will make the function negative.  So, the values less than -4 or greater than 3 will make the function positive.  Therefore, B is the correct answer.

Another way to do this problem is to graph the function and see where the graph is above the x-axis.  But, sometimes it is not easy to graph the function.

4 0
3 years ago
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