Hey there!!
Let's take the two numbers as 'x' and 'y'
The sum of these two numbers is 36.
... x + y = 36
Twice the first number minus the second is 6.
... 2x - y = 6
Now, we have two equations.
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... x + y = 36
... 2x - y = 6
Now, let's add both the equations.
... 3x = 42
Divide each side by 3.
... x = 42 / 3
... x = 14
<em>Hence, the first number is'14'. </em>
We know the sum of both the numbers is 36.
... x + y = 36
... 14 + y = 36
Subtract 14 on both sides.
... y = 36-14
... y = 22
<em>Hence, the second number is 22. </em>
<em>The numbers are 14 and 22. </em>
Hope my answer helps!!
Answer:
2
Step-by-step explanation:
19 minus 17 equals 2 .
The area of a square is A=s^2
We already have the area, so we solve for the side length:
169=s^2
13=s
The perimeter of a square is P=4s
We plug in 13 for s:
P=4(13)
P=52
So the perimeter of the square is 52 m^2
Hope this helped!
It’s choice D. Z, hope that helps :)
Answer:
x = 3.24, x = -1.24
Step-by-step explanation:
The standard form for a quadratic equation is 
. For your equation a = 1, b = -2, c = -4. The quadratic formula you will be using is 
.
Plug in a = 1, b = -2, and c = -4 into the formula.
We'll do the top part first:
Apply rule 
Apply exponent rule 
if 
is even
Multiply the numbers
Add
The prime factorization of 20 is 
20 divides by 2. <em>20 = 10 * 2</em>
<em />
<em />
10 divides by 2. <em>10 = 5 * 2</em>
<em />
<em />
2 & 5 are prime numbers so you don't need to factor them anymore
Apply radical rule ![\sqrt[n]{ab} =\sqrt[n]{a} \sqrt[n]{b}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%20%3D%5Csqrt%5Bn%5D%7Ba%7D%20%5Csqrt%5Bn%5D%7Bb%7D)
Apply radical rule ![\sqrt[n]{a^{n} } =a](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5E%7Bn%7D%20%7D%20%3Da)
; 
Because of the 
you have to separate the solutions so that one is positive and the other is negative.
Positive x:
Apply rule 
Multiply
Factor 
and rewrite it as 
. Factor out 2 because it is the common term. 
.
Divide 2 by 2
or 
(You'll probably have to use a calculator for the square root of 5)
^Repeating the process of positive x for negative x in order to get 
or 
