The value of x is 6, then the value of JK is 83, KL is 83 and JL is 83.
<h3>What is a triangle?</h3>
A triangle is a polygon that has three sides and three vertices. It is one of the basic figures in geometry.
An equilateral triangle is a shape with three sides that have an equal measure of side length.
The 3 sides of triangle JKL are equal to each other, therefore we can equate any of the two sides together to find the unknown variable called x.
So, JK = KL
JK = 13x + 5
KL + 17x - 19
13x + 5 = 17x - 19
Collect like terms;
17x - 13x = 19 + 5
4x = 24
x = 24/4
x = 6
We can now proceed to find the sides;
JK = 13x + 5
13 x 6 + 5
= 78 + 5
= 83
JK = 83
KL = 17x - 19
= 17 x 6 - 19
= 102 - 19
= 83
KL = 83
JL= 8x + 35
= 8 x 6 + 35
= 48 + 35
= 83
JL = 83
Hence, The value of x is 6, then the value of JK is 83, KL is 83 and JL is 83.
Learn more about the triangles;
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15 increased by a number x: 15 + x
a number x more than −14: x > -14
t<span>he difference of 7 and a number x: </span>7/x
Answer:
good luck bro these types of questions are so annoying im doing one now and is the question from mathswatch
Step-by-step explanation:
Answer:
Solution : (15, - 11)
Step-by-step explanation:
We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )
Row Echelon Form :

Step # 1 : Swap the first and second matrix rows,

Step # 2 : Cancel leading coefficient in row 2 through
,

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

As you can see our solution is x = 15, y = - 11 or (15, - 11).
Answer:
(x + 14)² + (y – 21/2)² = 1
Step-by-step explanation:
The equation of a circle can be written as seen below
(x – h)² + (y – k)² = r²
Where (h,k) is at the center and r = radius
We are given that the radius is 1
We are also given that the center is at (-14,21/2)
So we know that r = 1, h = -14, and k = 21/2
So to find the equation of the circle we simply substitute these values into the equation of a circle
Equation of a circle: (x – h)² + (y – k)² = r²
r = 1, h = -14, and k = 21/2
Substitute values
(x – (-14))² + (y – 21/2)² = 1²
1^2 = 1
The two negative signs before the 14 cancel out and it changes to + 14
The equation of a circle with a center at (-14,21/2) and a radius of 1 is (x + 14)² + (y – 21/2)² = 1