Answer:
I don’t know what 14 and 15 are but I know 16 is 6 ft
Step-by-step explanation:
Lets calculate how much she walks each weeks and then we multiply that by the number of weeks.
weekWalk = 5(2) + 3 = 10 + 3 = 13
therefore, she walks 13 miles per week, if she walked for 29 weeks then we have:
total walked =(29)(13) = 377
so she walked 377 miles total, so her original estimate is not reasonable
<span>f(x)=(<span><span>3x−1)/</span>2<----Starter </span></span>
y = (3x-1)/2
switch x and y
x = (3y-1)/2
2x = 3y-1
2x+1 =3y
y = 2/3*x + 1/3
f^(-1) (x) = 2/3*x + 1/3
f^(-1) (4) = 2/3*4 + 1/3 = 8/3 + 1/3 = 9/3 = 3
I hope this helps. ;)
Answer:
9
Step-by-step explanation:
Because line PQ and line PR are both tangent to circle S, it would result in them being equal. So first to find PQ, you would have to find the value of variable x. To do this you would set both equations equal to each other and solve
6x - 9 = 2x + 3
-2x -2x
4x - 9 = 3
+9 +9
4x = 12
x = 3
Now you have to substitute 3 back into the equation for PQ:
6(3) - 9
18 - 9
= 9
Resulting in PQ = 9, to check your answer, take the value of x (3) and substitute it into PR and you should also get 9
2(3) + 3
6 + 3
= 9
The percentage of the semicircle shaded section is approximately 23,606 %.
The percentage of the area of the semicircle is equal to the ratio of the semicircle area minus the half-cross area to the semicircle area. In other words, we have the following expression:
(1)
Where:
- Area of the half cross, in square centimeters.
- Area of the semicircle, in square centimeters.
- Percentage of the shaded section of the semicircle.
And the percentage of the shaded section is:
![r = \left[1-\frac{4 \cdot (2\,cm)^{2}+4\cdot \left(\frac{1}{2} \right)\cdot (2\,cm)^{2}}{0.5\cdot \pi\cdot (16\,cm^{2}+4\,cm^{2})} \right]\times 100](https://tex.z-dn.net/?f=r%20%3D%20%5Cleft%5B1-%5Cfrac%7B4%20%5Ccdot%20%282%5C%2Ccm%29%5E%7B2%7D%2B4%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5Ccdot%20%282%5C%2Ccm%29%5E%7B2%7D%7D%7B0.5%5Ccdot%20%5Cpi%5Ccdot%20%2816%5C%2Ccm%5E%7B2%7D%2B4%5C%2Ccm%5E%7B2%7D%29%7D%20%5Cright%5D%5Ctimes%20100)
The percentage of the semicircle shaded section is approximately 23,606 %.
We kindly invite to check this question on percentages: brainly.com/question/15469506
