Answer:
Step-by-step explanation:
the figure is divided into a cube and a cuboid
for the cube,
volume = a^3
= 6^3 = 216 cm^3
for the cuboid,
volume = l × b × h
= 13 × 6 × 6
=468 cm^3
so total volume = volume of the cube + volume of the cuboid
= 216 + 468
=684 cm^3
hope this helps
plz mark it as brainliest!!!!!!!!
Answer:
t = 5.46 secs and t = 2.42 secs
Step-by-step explanation:
The distance of the ball from the ground is given by:
![s(t) = -16t^2 + 126t](https://tex.z-dn.net/?f=s%28t%29%20%3D%20-16t%5E2%20%2B%20126t)
When the ball is at a distance 211 ft from the ground, that means s(t) = 211 ft. we need to find t:
![211 = -16t^2 + 126t](https://tex.z-dn.net/?f=211%20%3D%20-16t%5E2%20%2B%20126t)
![=> 16t^2 -126t + 211 = 0](https://tex.z-dn.net/?f=%3D%3E%2016t%5E2%20-126t%20%2B%20211%20%3D%200)
Using the quadratic formula:
and ![t = \frac{-b - \sqrt{b^2 - 4ac} }{2a}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-b%20-%20%5Csqrt%7Bb%5E2%20-%204ac%7D%20%7D%7B2a%7D)
where a = 16, b = 126 and c = 211
Hence:
and ![t = \frac{-(-126) - \sqrt{(-126)^2 - 4(16)(211)} }{2(16)}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-%28-126%29%20-%20%5Csqrt%7B%28-126%29%5E2%20-%204%2816%29%28211%29%7D%20%7D%7B2%2816%29%7D)
and ![t = \frac{126 - \sqrt{15876 - 13504} }{32}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B126%20-%20%5Csqrt%7B15876%20-%2013504%7D%20%7D%7B32%7D)
and ![t = \frac{126 - \sqrt{2372} }{32}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B126%20-%20%5Csqrt%7B2372%7D%20%7D%7B32%7D)
and ![t = \frac{126 - 48.703 }{32}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B126%20-%2048.703%20%7D%7B32%7D)
and ![t = \frac{77.297}{32}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B77.297%7D%7B32%7D)
=> t = 5.46 secs and t = 2.42 secs
This means that the ball will be at 211 feet at two different times, first, after 2.42 seconds and then after 5.46 seconds.
This makes sense because the ball is projected upwards, which means that when going up, it attains 211 feet and also, when coming back down, it also attains 211 feet.
Answer:
p = (2k + 1)i + j
q = 25i + (k + 4)j
p . q = 11
Find the dot product of p and q
(2k + 1)(25) + (k + 4) = 11
50k + 25 + k + 4 = 11
51k = -18
k = - 18/51 = –6/17.✅
b) p.q = |p||q|cos∅
p= (2k + 1)i + j = [2(-6/17) + 1)i + j ]= 5/17i + j
|p| = √ (5/17)² + 1²
|p| = √314/ 17 ( The square root doesn't cover the 17) = 1.0424
q = 25i + (k + 4)j = 25i + ( -6/17 + 4)j
q = 25i + 62/17 j
|q| = √ 25² + (62/17)²
= 25.2646
Now applying them to the formula above
We already have p.q = 11 from the question.
11 = (1.0424)(25.2646)Cos∅
11 = 26.3358Cos∅
Cos∅ = 11/26.3358
Cos∅ = 0.4177
∅ = 65.31°.
This should be it.
Hope it helps
We need a picture to understand the question :)
Given that the width of the rectangular piccture is:
![w=\frac{1}{2}l](https://tex.z-dn.net/?f=w%3D%5Cfrac%7B1%7D%7B2%7Dl)
Where "l" is the length of the picture.
You know that the perimeter is:
![P=72\text{ }in](https://tex.z-dn.net/?f=P%3D72%5Ctext%7B%20%7Din)
The formula for calculating the perimeter of a rectangle is:
![P=2l+2w](https://tex.z-dn.net/?f=P%3D2l%2B2w)
Where "l" is the length and "w" is the width.
Therefore, you can substitute the perimeter and the width into the formula, in order to find "l":
![72=2l+2(\frac{1}{2}l)](https://tex.z-dn.net/?f=72%3D2l%2B2%28%5Cfrac%7B1%7D%7B2%7Dl%29)
Now you can solve for "l":
![72=2l+\frac{2}{2}l](https://tex.z-dn.net/?f=72%3D2l%2B%5Cfrac%7B2%7D%7B2%7Dl)
![72=2l+l](https://tex.z-dn.net/?f=72%3D2l%2Bl)
![72=3l](https://tex.z-dn.net/?f=72%3D3l)
![\frac{72}{3}=l](https://tex.z-dn.net/?f=%5Cfrac%7B72%7D%7B3%7D%3Dl)
![l=24\text{ }in](https://tex.z-dn.net/?f=l%3D24%5Ctext%7B%20%7Din)
Hence, the answer is: