Answer:
-30-45k
Step-by-step explanation:
-5*6=-30
-5*9k=-45k
I'm not sure if you're asking for a number in place of Q, but if so then:
Anything between 1 and 2.
12 * 1.5 = 18.
Answer:
1.25 cm
Step-by-step explanation:
Draw a radius line from the center to one of the top corners. This forms a right triangle where the hypotenuse is r, the width is 1, and the height is 2−r. Using Pythagorean theorem:
(2−r)² + 1² = r²
4 − 4r + r² + 1 = r²
5 − 4r = 0
r = 5/4
r = 1.25
we can pretty much split the middle part into two trapezoids. Check the picture below.
so we really have one trapezoid and one square, each twice, so simply let's get the area of the trapezoid and sum it up with the area of the square, twice, and that's the area of the shape.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{\textit{parallel sides}}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=3\\ b=7 \end{cases}\implies A=\cfrac{5(3+7)}{2}\implies A=25 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{sum of areas}}{[25+(3\cdot 3)]}\cdot \stackrel{twice}{2}\implies [34]2\implies \underset{in^2}{68}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7B%5Ctextit%7Bparallel%20sides%7D%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D5%5C%5C%20a%3D3%5C%5C%20b%3D7%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B5%283%2B7%29%7D%7B2%7D%5Cimplies%20A%3D25%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20areas%7D%7D%7B%5B25%2B%283%5Ccdot%203%29%5D%7D%5Ccdot%20%5Cstackrel%7Btwice%7D%7B2%7D%5Cimplies%20%5B34%5D2%5Cimplies%20%5Cunderset%7Bin%5E2%7D%7B68%7D)
Answer:
6a) i- 2hrs 36mins ii- 3hrs 12mins
b) car A≈ 76.9km/h car B≈ 62.5km/h
c)------
7a) 35km
b) car A=75km car B=60km
c) 30km
d) car A≈36mins car B≈48mins
Step-by-step explanation:
6a) Using the graph follow the lines until they finish then go downwards until you get to the x-axis. The x-axis is going up by 12mins for each square.
b) Using the answer from a, you divide 200km by the time.
For car A 2hrs 36mins becomes 2.6 because 36mins/60mins=0.6
∴ car A: 200/2.6≈ 76.92km/h
For car B 3hrs 12mins becomes 3.2 because 12mins/60mins=0.2
∴ car B: 200/3.2≈ 62.5km/h
7a) Using the graph go down from where the line of car A finished to meet car B. The y-axis is going up by 5km for each square.
b) Starting from the x-axis at 1 hour go upwards to see where you meet the car B line (60km) and car A line(75km). (sorry if that does not really make sense).
c) Difference from car A line to car B:
155km-125km=30km
d) Going across from 50km meet car A line and go down to see it has been travelling for approx. 36mins. Then continue across to car B line, go down to see it reached 50km at approx. 48mins.
Hope this helps.
