We can factor by grouping. To do so, we multiply the leading coefficient with the constant at the end. In other words, a times c (ax^2 + bx + c).
15*-4 = -60
Now we need to split the b term into two pieces that multiply to -60 and add to 4.
-6 and 10 will work.
Now group one part of b with the 15x^2 and the other part with -4.
(15x^2 + 10x) + (-6x - 4)
Now factor both terms.
5x(3x+2) - 2(3x+2)
3x+2 is one of our factors and 5x-2 is the other.
(3x+2)(5x-2)=0
Now just find the zeros.
3x+2 = 0
3x = -2
x = -2/3
And
5x-2 = 0
5x = 2
x = 2/5
So the answer is x = -2/3 and x = 2/5
6(x+1)=24
6x+6=24
6x=24-6
6x=18
6x/6=18/6
x=3
From the box plot, it can be seen that for grade 7 students,
The least value is 72 and the highest value is 91. The lower and the upper quartiles are 78 and 88 respectively while the median is 84.
Thus, interquatile range of <span>the resting pulse rate of grade 7 students is upper quatile - lower quartle = 88 - 78 = 10
</span>Similarly, from the box plot, it can be seen that for grade 8 students,
The
least value is 76 and the highest value is 97. The lower and the upper
quartiles are 85 and 94 respectively while the median is 89.
Thus, interquatile range of the resting pulse rate of grade 8 students is upper quatile - lower quartle = 94 - 85 = 9
The difference of the medians <span>of the resting pulse rate of grade 7 students and grade 8 students is 89 - 84 = 5
Therefore, t</span><span>he difference of the medians is about half of the interquartile range of either data set.</span>
Answer:
Line sw has an underfined slope
Line tv has a slope of 0
Line rs and tv are parallel
Step-by-step explanation: I got it right on my quiz
5) slope is y=mx+b
3) cross the y-axis
1) perpedicular to the y-axis
4) parallel
sorry it's in a random order and idk #2