Answer: 615.4mm cube
Step-by-step explanation:
The volume of a cylinder is given as πr^2h
where r= radius
h = height
π = 3.14
From the question, diameter is 14mm and height is 4mm.
To get the radius, we divide the diameter by 2 which will give:
= 14mm/2 = 7mm
Volume= πr^2h
= 3.14 × 7^2 × 4
= 3.14 × 49 × 4
= 615.4mm cube
The volume of the cylindrical barrel is 615.4mm cube.
Answer:
<em>The solution of the system is: 
</em>
Step-by-step explanation:
The given system of equations is.......
So, the augmented matrix will be: ![\left[\begin{array}{cccc}-1&-3&|&-17\\2&-6&|&-26\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-1%26-3%26%7C%26-17%5C%5C2%26-6%26%7C%26-26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now, we will transform the augmented matrix to the reduced row echelon form using row operations.
<u>Row operation 1 :</u> Multiply the 1st row by -1. So..........
![\left[\begin{array}{cccc}1&3&|&17\\2&-6&|&-26\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C2%26-6%26%7C%26-26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 2:</u> Add -2 times the 1st row to the 2nd row. So.......
![\left[\begin{array}{cccc}1&3&|&17\\0&-12&|&-60\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C0%26-12%26%7C%26-60%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 3:</u> Multiply the 2nd row by 
. So.......
![\left[\begin{array}{cccc}1&3&|&17\\0&1&|&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C0%261%26%7C%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 4:</u> Add -3 times the 2nd row to the 1st row. So........
![\left[\begin{array}{cccc}1&0&|&2\\0&1&|&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%7C%262%5C%5C0%261%26%7C%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now, from this reduced row echelon form of the augmented matrix, we can get that 
and 
So, the solution of the system is:
Answer:
A. 6 and 30
B. 8 and 28
Step-by-step explanation:
we know that the area of the three squares must satisfy the Pythagorean Theorem so c² = a² + b²
where
c² is the area of the largest square
a² and b² are the areas of the smaller squares
now.. 36² = a² + b2
The sum of the areas of the smaller squares must be equal to 36
therefore
6 and 30 >>>> could be the areas of the smaller squares (6 + 30 = 36)
8 and 28 >>>> could be the areas of the smaller squares (8 + 28 = 36)
hope it clears your mind.
Please mark as the brainliest
The way to answer this problem given different points and different equations in the choices is to do substitution and trial and error. In this regard, we substitute for example 4 to x in the equations and check which of them yields 89. The answer is D.
Answer:
i only know acdc
Step-by-step explanation:
