Answer:
75.89
Step-by-step explanation:
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Answer:
r = 5mm
Step-by-step explanation:
When two shapes are similar, They are the same shape but with 1 or more basic transformations(dilations, reflections, rotations, and translations.). All sides on the new shape (green) are similar to the original shape (blue), meaning they were all dilated by the same scale factor.
First, we need to find our scale factor. To find our scale factor, take 10 and divide it by 4 (These sides are similar and we have established values so we can be sure our scale factor is correct by using these.).
10/4 = 2.5
So, our scale factor is 2.5.
Now, to solve for r, we need to find it's similar side (in this case it's 2 mm.).
Now, simply multiply the original size by our scale factor.
2 * 2.5 = 5.
So, r = 5 mm (Don't forget units, teachers hate that.)
18)
It says the scale factor is 1:25000, so for every one part there is actually 25000. So for every centimeter, there is actually 25000 cm. See that we multiplied by 25000.
So for 6.8cm, multiply by 25000 to get 170000cm.
The real distance represented is 170000cm.
19) Just work out one step at a time using order of operations.
Do parentheses first. 7^2 = 49, and 3^3 is 3*3*3 which is 27.
Now you have [8(49 - 27) ]/ sqrt121.
Finish simplifying the parentheses. 49 - 27 is 22.
[8(22)] / sqrt121.
Now multiply 8 times 22, which is 176.
176 / sqrt 121
The square root of 121 is 11.
176/11.
= 16
Answer:
The zeros of this function are x=(1,7)
Step-by-step explanation:
Check the picture below.
well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.
bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.
![\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}](https://tex.z-dn.net/?f=%5Cbf%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-3%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bslope%20of%20AB%7D%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7B%5Cunderline%7Bnegative%20reciprocal%7D%20and%20slope%20of%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D)
so, it passes through the midpoint of AB,
so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)
