Answer: -2
Different in Squares: a^2 −b^2 =(a+b)(a−b)
Rad 7^2-3^2 —> Rad 7^2=7 and -3^2= -9
—> 7-9= -2
Answer:
As x ⇒-∞, P(x) ⇒ -∞
As x ⇒ ∞, P(x) ⇒ ∞
Step-by-step explanation:
To find left hand end behavior, plug in negative infinity into the function and evaluate...
P(x) = 3(-∞) = -3(∞) = -∞
The 'y' values of the function decrease towards negative infinity as the 'x' values approach negative infinity
P(x) = 3(∞) = ∞
The 'y' values of the function increase towards positive infinity as the 'x' values approach positive infinity
perimeter = 2width + 2 length
102 = 2(x+9) + 2x
102 = 2x + 18 +2x
102 = 4x +18
84 = 4x
21 = x
Length = x = 21 inches
Width = x + 9 = 21 + 9 = 30 inches
Answer:
The value of x is equal to -2/3.
Step-by-step explanation:
In the problem it says that y equals 2x, and that 3x - 3y = 2.
The first step is to substitute 2x into y.
3x - 3(2x) = 2.
The next step is to use distributive property.
-3(2x) = -6x.
Now we need to add like terms.
3x - 6x = -3x.
Which gives us the equation -3x = 2.
The final step is to divide on both sides to get the value of x.
-3x/-3 = 2/-3.
x = -2/3.
So as you can see, x is equal to -2/3.
We can also check to make sure by redoing the problem, but substituting the value of x.
3(-2/3) - 3(2 * -2/3) = 2.
-2 - 3 * -4/3 = 2.
-2 + 4 = 2.
2 = 2.
The value of x is indeed -2/3.
Answer:
○ C
Explanation:
Accourding to one of the circle equations, 
the centre of the circle is represented by 
Moreover, all negative symbols give you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must pay cloce attention to which term gets which symbol. Another thing you need to know is that the radius will ALWAYS be squared, so no matter how your equation comes about, make sure that the radius is squared. Now, in case you did not know how to define the radius, you can choose between either method:
Pythagorean Theorem
Sinse we are dealing with <em>length</em>, we only desire the NON-NEGATIVE root.
Distanse Equation
![\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = d \\ \\ \sqrt{[-5 - 3]^2 + [3 + 3]^2} = r \hookrightarrow \sqrt{[-8]^2 + 6^2} = r \hookrightarrow \sqrt{64 + 36} = r; \sqrt{100} = r \\ \\ \boxed{10 = r}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E2%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E2%7D%20%3D%20d%20%5C%5C%20%5C%5C%20%5Csqrt%7B%5B-5%20-%203%5D%5E2%20%2B%20%5B3%20%2B%203%5D%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B%5B-8%5D%5E2%20%2B%206%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B64%20%2B%2036%7D%20%3D%20r%3B%20%5Csqrt%7B100%7D%20%3D%20r%20%5C%5C%20%5C%5C%20%5Cboxed%7B10%20%3D%20r%7D)
Sinse we are dealing with <em>distanse</em>, we only desire the NON-NEGATIVE root.
I am joyous to assist you at any time.
