Applying the Pythagorean theorem, the measure of the other leg, to the nearest tenth, is: 2.6 cm.
<h3>What is the
Pythagorean Theorem?</h3>
In a right triangle, where the lengths of its two legs are a and b, and the hypotenuse length is c, the Pythagorean theorem states that: c² = a² + b².
Given the following parameters of the right triangle:
- c = 4 cm
- a = 3 cm
- b = ? (other leg)
Apply the Pythagorean theorem:
b = √(4² - 3²)
b = 2.6 cm (nearest tenth)
Thus, applying the Pythagorean theorem, the measure of the other leg, to the nearest tenth, is: 2.6 cm.
Learn more about Pythagorean theorem on:
brainly.com/question/654982
Answer:
3/5
Step-by-step explanation:
using y=mx + b= y2-y1/x2- x1
Answer:
The solution to the system of equations is
![\begin{gathered} x=\frac{179}{13} \\ \\ y=-\frac{279}{39} \\ \\ z=-\frac{48}{13} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B179%7D%7B13%7D%20%5C%5C%20%20%5C%5C%20y%3D-%5Cfrac%7B279%7D%7B39%7D%20%5C%5C%20%20%5C%5C%20z%3D-%5Cfrac%7B48%7D%7B13%7D%20%5Cend%7Bgathered%7D)
Explanation:
Giving the system of equations:
![\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%2B3y-z%3D-4%5Cldots%5Cldots%5Cldots%5Cldots%5Cldots%5Cldots..........%5Cldots%5Cldots%5Cldots%5Cldots.%5Cldots%5Ctext%7B.%7D%5Cmathrm%7B%7D%281%29%20%5C%5C%202x-y%2B2z%3D13%5Cldots%5Cldots...%5Cldots%5Cldots%5Cldots%5Cldots..%5Cldots..%5Cldots%5Cldots%5Cldots%5Cldots%5Cldots.%282%29%20%5C%5C%203x-2y-z%3D-9%5Cldots%5Cldots%5Cldots.%5Cldots%5Cldots%5Cldots%5Cldots....%5Cldots%5Cldots.%5Cldots%5Cldots%5Cldots%5Ctext%7B.%7D%5Cmathrm%7B%7D%283%29%20%5Cend%7Bgathered%7D)
To solve this, we need to first of all eliminate one variable from any two of the equations.
Subtracting (2) from twice of (1), we have:
![5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)](https://tex.z-dn.net/?f=5y-4z%3D-21%5Cldots%5Cldots%5Cldots%5Cldots%5Cldots.%5Cldots.%5Cldots..%5Cldots..%5Cldots%5Cldots.%5Cldots..%5Cldots%5Ctext%7B...%7D%5Cmathrm%7B%7D%284%29)
Subtracting (3) from 3 times (1), we have
![3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)](https://tex.z-dn.net/?f=3y-5z%3D-3%5Cldots%5Cldots...%5Cldots%5Cldots..%5Cldots%5Cldots%5Cldots%5Cldots%5Cldots.%5Cldots%5Cldots%5Cldots%5Cldots%5Cldots..%5Cldots%5Cldots%285%29)
From (4) and (5), we can solve for y and z.
Subtract 5 times (5) from 3 times (4)
![\begin{gathered} 13z=-48 \\ \\ z=-\frac{48}{13} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%2013z%3D-48%20%5C%5C%20%20%5C%5C%20z%3D-%5Cfrac%7B48%7D%7B13%7D%20%5Cend%7Bgathered%7D)
Using the value of z obtained in (5), we have
![\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\ \\ 3y+\frac{240}{13}=-3 \\ \\ 3y=-3-\frac{240}{13} \\ \\ 3y=-\frac{279}{13} \\ \\ y=-\frac{279}{39} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%203y-5%28-%5Cfrac%7B48%7D%7B13%7D%29%3D-3%20%5C%5C%20%20%5C%5C%203y%2B%5Cfrac%7B240%7D%7B13%7D%3D-3%20%5C%5C%20%20%5C%5C%203y%3D-3-%5Cfrac%7B240%7D%7B13%7D%20%5C%5C%20%20%5C%5C%203y%3D-%5Cfrac%7B279%7D%7B13%7D%20%5C%5C%20%20%5C%5C%20y%3D-%5Cfrac%7B279%7D%7B39%7D%20%5Cend%7Bgathered%7D)
Using the values obtained for y and z in (1), we have
Answer:
I am not sure
Step-by-step explanation:
The base of a cone is something circular, like a circle or an ellipse.
The rest of the options (square, triangle, and pentagon) are all based of pyramids, not cones. Cones have to have a circular base because otherwise they'd be pyramids.