Answer:
the first one will be a <u>positive correlation</u>
the answer for the second Q will be 16
Step-by-step explanation:
have a nice day
f<span>(x)</span>=<span>x^2</span>−<span>6
</span>Replace <span>f<span>(x)</span></span> with <span>yy</span>.
<span>y=<span>x^2</span>−<span>6
</span></span>Interchange the variables.
<span>x=<span>y2</span>−6
</span>Solve for <span>yy</span><span>.
</span>
Move <span><span>−6</span><span>-6</span></span> to the right side of the equation by subtracting <span><span>−6</span><span>-6</span></span> from both sides of the equation.<span><span><span>y2</span>=6+x</span><span><span>y2</span>=6+x</span></span>Take the <span><span>square</span><span>square</span></span> root of both sides of the <span><span>equation</span><span>equation</span></span> to eliminate the exponent on the left side.<span><span>y=±<span>√<span>6+x</span></span></span><span>y=±<span>6+x
</span></span></span>The complete solution is the result of both the positive and negative portions of the solution.
Tap for more steps...
<span>y=<span>√<span>6+x</span></span>,−<span>√<span>6+x</span></span></span>
Solve for y<span> and replace with </span><span><span>f^<span>−1</span></span><span>(x).
</span></span>
<span>Answer is f<span>−1</span></span><span>(x)</span>=<span>√<span>6+x</span></span>,−<span>√<span>6+<span>x</span></span></span>
Answer:
see explanation
Step-by-step explanation:
A recursive rule allows us to find the term in a sequence from the previous term.
From the given geometric sequence find r the common ratio
r = 
= 
= - 4
Hence recursive rule is
= - 4
( a₁ = 
)
Answer:
f(1/2) = -2
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
f(x) = 8x - 6
f(1/2) is x = 1/2
<u>Step 2: Evaluate</u>
- Substitute in <em>x</em>: f(1/2) = 8(1/2) - 6
- Multiply: f(1/2) = 4 - 6
- Subtract: f(1/2) = -2
Answer:
ASA
Step-by-step explanation:
∠BAC ≈ ∠CAD
AC = AC
∠BCA ≈ ∠ACD
