The proof is given below. Please go through it.
Step-by-step explanation:
To solve Δ ABC ≅ Δ DBC
From Δ ABC and Δ DBC
AB = BD (given)
AC = CD (given)
BC is common side
By SSS condition Δ ABC ≅ Δ DBC ( proved)
To solve Δ EHF ≅ Δ GHF
Δ EHF and Δ GHF
EH = HG ( given)
∠ EFH = ∠ GFH ( each angle is 90°)
HF is common side
By RHS condition
Δ EHF ≅ Δ GHF
Answer:
765
Step-by-step explanation:
Answer:is (4x-5)
Step-by-step explanation:
Answer:
Step-by-step explanation:
The easiest way to do this is by using synthetic division (I really hope you know what I'm talking about!)
If the factor is x + 2 = 0, then the solution or the root is x = -2. We will put -2 outside the "box" and the coefficients from the terms of the polynomial inside the "box":
-2| 1 0 0 8
Bring down the first number and multiply it by the -2. Put the product up under the first 0 and add:
-2| 1 0 0 8
-2
1 -2
Now multiply the -2 by the -2, put the product up under the next 0 and add:
-2| 1 0 0 8
-2 4
1 -2 4
Now multiply the 4 by the -2, put the product up under the 8 and add:
-2| 1 0 0 8
-2 4 -8
1 -2 4 0
Those bold numbers are the coefficients to the quotient, which is another polynomial, but is one degree less than the one you started with:
x² - 2x + 4 = 0
Since there is no remainder, then x = -2 is a zero of x³ + 8, and (x + 2) is a factor.
There is a whole world of stuff to learn about quadratics!!
Answer:
Step-by-step explanation:
<u>The Inverse of a Function</u>
Given a function f(x), the inverse of f, called 
is a function that satisfies:
The domain of f becomes the range of its inverse and vice-versa.
The procedure to find the inverse of the function is:
* Write the function as a two-variable equation:
* Solve the equation for x.
* Swap the variables
We are given the function:
Defined in the interval x ≥ -3
Note f is always negative, thus its range is f(x) ≤ 0
Now we find the inverse. Call y=f(x):
Multiplying by 2:
Squaring both sides:
Subtracting 3:
Swapping variables:
Thus:
The domain of the inverse is the range of f, thus x is restricted to
x ≤ 0
