The question is an illustration of ratios and proportions
<em>The Bospolder fox is 28 inches tall at the shoulder,</em>
Represent the ratio as:
![Scale = Shoulder : Tail](https://tex.z-dn.net/?f=Scale%20%3D%20Shoulder%20%3A%20Tail)
For the red fox, we have:
![Red = 20in : 55in](https://tex.z-dn.net/?f=Red%20%3D%2020in%20%3A%2055in)
For the Bospolder fox, we have:
![Bospolder =Shoulder : Tail](https://tex.z-dn.net/?f=Bospolder%20%3DShoulder%20%3A%20Tail)
To calculate the length of the shoulder of the Bospolder fox, we need its length at the tail.
<em>Since it is not given, I will make an assumption.</em>
Assume, the Bospolder fox is 77in tall at the tail.
The ratio becomes
![Bospolder =Shoulder : 77in](https://tex.z-dn.net/?f=Bospolder%20%3DShoulder%20%3A%2077in)
Both animals are proportional; this means that they have the same ratio.
So, we have:
![Shoulder : 77in = 20in : 55in](https://tex.z-dn.net/?f=Shoulder%20%3A%2077in%20%3D%2020in%20%3A%2055in)
Express as fraction
![\frac{Shoulder }{ 77in }= \frac{20in }{55in}](https://tex.z-dn.net/?f=%5Cfrac%7BShoulder%20%7D%7B%2077in%20%7D%3D%20%5Cfrac%7B20in%20%7D%7B55in%7D)
Multiply both sides by 77in
![Shoulder = \frac{20in }{55in} \times 77in](https://tex.z-dn.net/?f=Shoulder%20%3D%20%5Cfrac%7B20in%20%7D%7B55in%7D%20%5Ctimes%2077in)
![Shoulder = \frac{20in \times 77}{55}](https://tex.z-dn.net/?f=Shoulder%20%3D%20%5Cfrac%7B20in%20%5Ctimes%2077%7D%7B55%7D)
![Shoulder = 28in](https://tex.z-dn.net/?f=Shoulder%20%3D%2028in)
Using the assumed length, the Bospolder fox is 28 inches tall at the shoulder,
Read more about ratios and proportions at:
brainly.com/question/13114933
Answer:
<em>The time it takes the ball to hit the ground is 3.05secs</em>
step - by - step explanation:
<em>In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05 </em>
two intersecting lines can only have 1 solution.
Answer:
Step-by-step explanation:
the shape is made up of 2 circles and one box add the 3 areas up
circle area = ![\pi](https://tex.z-dn.net/?f=%5Cpi)
![r^{2}](https://tex.z-dn.net/?f=r%5E%7B2%7D)
3.14*
= 50.24
50.24*2 =100.48
box, box, pit now :P ooh I was thinking formula one racing, ( i've been watching that lately) anyway
box = 8*8 = 64
total = 164.48 is that an option? hmm, yes the bottom two choices.. now the perimeter . the perimeter is made up of the 2 circles
the circumference of a circle is 2
r
2*
*r
2*
*4 =25.1327 that's one whole circle , multiplied by 2 , I can see , by math in my head that's about 50.26
sooo
the 3rd choice looks good :)