the picture in the attached figure
we know that
area of the triangle is equal to
in this problem
A=47 in²
b=x
h=x
so
therefore
the answer is
x=9.7 in
9514 1404 393
Answer:
0
Step-by-step explanation:
The forces are equal and opposite, so the net force is zero.
5N -5N = 0
<h3><u>
Answer:</u></h3>
<h3><u>
Step-by-step explanation:</u></h3>
We know that:
- <u>Trapezoid ABCD has one side that is 25 unit. </u>
- <u>The other 3 sides (AB, BC, and CD) are 12 units. </u>
- <u>EF = 6 units</u>
- <u>EH = 12.5</u>
- <u>Trapezoid ABCD is congruent to Trapezoid EFGH</u>
<em>If the 3 sides of the trapezoid ABCD are the same sides, then side EF, FG, GH must be of the same length because of congruence. The value of FG and GH must be the same length as EF. We can clearly see in the picture that EF is 6 units. Hence, EF is 6 units, FG is 6 units, and GH is 6 units. The work of the perimeter is shown below.</em>
<u>Work</u>
- => 6(3) + 12.5
- => 18 + 12.5
- => <u>30.5 units</u>
Hence, the perimeter of EFGH is 30.5 units.
100 times 17 bc u have to count the shdded parts
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>
