It will be clearly B.38mm3
Consider, in ΔRPQ,
RP = R (Radius of larger circle)
PQ = r (radius of smaller circle)
We have to find, RQ, by Pythagoras theorem,
RP² = PQ²+RQ²
R² = r²+RQ²
RQ² = R²-r²
RQ = √(R²-r²
Now, as RQ & QS both are tangents of the smaller circle, their lengths must be equal. so, RS = 2 × RQ
RS = 2√(R²-r²)
The <span>given the piecewise function is :
</span>
![f(x) = \[ \begin{cases} 2x & x \ \textless \ 1 \\ 5 & x=1 \\ x^2 & x\ \textgreater \ 1 \end{cases} \]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5C%5B%20%5Cbegin%7Bcases%7D%20%0A%20%20%20%20%20%202x%20%26%20x%20%5C%20%5Ctextless%20%5C%20%201%20%5C%5C%0A%20%20%20%20%20%205%20%26%20x%3D1%20%5C%5C%0A%20%20%20%20%20%20x%5E2%20%26%20x%5C%20%5Ctextgreater%20%5C%201%20%0A%20%20%20%5Cend%7Bcases%7D%0A%5C%5D)
To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25
To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4
To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4
To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
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So, the statements which are true:<span>

</span><span>
</span>
Answer: do not know just do it
Step-by-step explanation:
Answer:
No its not
Step-by-step explanation:
When 42(x + 14y) is simplified:
42(x + 14y)
Expand and remove the brackets. Remember brackets mean to multiply.
42*x
42x
42*14y
588y
42x+588y
Therefore 42x-56 is NOT EQUIVALENT to 42(x + 14y) but is equivalent to 42+588y