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Nady [450]
2 years ago
10

Please help me witht this

Mathematics
2 answers:
iogann1982 [59]2 years ago
7 0

Answer:

B,D

Step-by-step explanation:

the independent variable is what is being changed or manipulated and he is changing how many plays he gets.  the amount of yards he gets is a result of the amount of plays so thats the dependent variable.  

Arisa [49]2 years ago
6 0

Answer: B and D

Step-by-step explanation:

The independent variable does not rely on any other factors and can be by itself. The dependent variable relies on the independent variable (the number of times he gets the ball directly affects how many yards he would gain).

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a class voted for either rollerblading, swimming, or biking as their favorite summer activity. if rollerblading got 21% percent
Sauron [17]

Answer:  39% voted for biking

Step-by-step explanation:

Add the other two percentages and subtract from 100

40 + 21 = 61

100 - 61 = 39

8 0
3 years ago
Can someone teach me how to balance algebra equations, I just don't get how to do it. Like with fractions?
natka813 [3]

Answer:

<h3>Move all the x terms to one side. Use inverse operations and add 1 5 x 15x 15x to both sides to keep the equation balanced. Solve by working backwards from the order of operations. This means we need to undo the −2 first by adding 2 to both sides of the equation to keep it balanced.</h3>

4 0
2 years ago
In 1980, the number of African elephants was 1,200,000 but that number was decreasing 5.25% a year.
algol13
<span>Answer is about 12.85.</span>
3 0
3 years ago
ASAP
Pie

Answer:

36/-4

Step-by-step explanation:

A positive number divided by a negative number will<em> always </em> be negative, and a negative number divided by a positive number will <em>always </em>be negative!

Hope this helps!

6 0
3 years ago
Read 2 more answers
Choose the point-slope form of the equation below that represents the line that passes through the points (−3, 2) and (2, 1).. .
aleksandrvk [35]
The\ slope-point\ form:y-y_1=m(x-x_1)\\m=\dfrac{y_2-y_1}{x_2-x_1}\\---------------------\\(-3;\ 2)\to x_1=-3\ and\ y_1=2\\(2;\ 1)\to x_2=2\ and\ y_2=1\\\\m=\dfrac{1-2}{2-(-3)}=\dfrac{-1}{5}=-\dfrac{1}{5}\\\\for\ (-3;\ 2)\\y-2=-\dfrac{1}{5}(x-(-3))\to y-2=-\dfrac{1}{5}(x+3)\\\\for\ (2;\ 1)\\y-1=-\dfrac{1}{5}(x+2)
6 0
3 years ago
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