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Let the side of the square be $s$. The perimeter of the square is $P = 4s$. We are given that the rate of decrease of the side is $\frac{ds}{dt} = -1.5~cm/s$. We need to find the rate of decrease of its perimeter, which is $\frac{dP}{dt}$. Differentiate the perimeter equation with respect to time $t$: $\frac{dP}{dt} = \frac{d}{dt}(4s)$ $\frac{dP}{dt} = 4 \frac{ds}{dt}$ Substitute the given value of $\frac{ds}{dt}$: $\frac{dP}{dt} = 4 (-1.5~cm/s)$ $\frac{dP}{dt} = -6~cm/s$ The rate of decrease of its perimeter is $6~cm/s$. The final answer is $\boxed{6~cm/s}$.

The solution proceeds as follows: **Step 1: Find the derivative of the function.** The derivative of \(f(x) = kx - \sin~x\) is \(f'(x) = k - \cos~x\). **Step 2: Determine the condition for a strictly increasing function.** For a function to be strictly increasing, its derivative must be strictly positive, i.e., \(f'(x) > 0\). **Step 3: Apply the condition to the derivative.** We need \(k - \cos~x > 0\) for all \(x\). This means \(k > \cos~x\) for all \(x\). **Step 4: Find the maximum value of $\cos~x$.** The maximum value of $\cos~x$ is 1. **Step 5: Determine the condition for $k$.** For \(k > \cos~x\) to hold for all \(x\), \(k\) must be greater than the maximum value of $\cos~x$. Therefore, \(k > 1\). **Final Answer:** The function \(f(x)=kx-\sin~x\) is strictly increasing for \(k\gt1\). The final answer is $\boxed{k\gt1}$.

To determine the correct statement about the function $f(x) = -2x^8$, we can analyze its first derivative to find critical points and intervals of increase/decrease. 1. **Calculate the first derivative:** $$f'(x) = \frac{d}{dx}(-2x^8) = -16x^7$$ 2. **Find critical points:** Set $f'(x) = 0$: $$-16x^7 = 0 \implies x^7 = 0 \implies x = 0$$ The only critical point is $x=0$. 3. **Analyze the sign of $f'(x)$:** * For $x < 0$, $x^7 < 0$, so $f'(x) = -16(\text{negative number}) > 0$. This means $f(x)$ is increasing for $x < 0$. * For $x > 0$, $x^7 > 0$, so $f'(x) = -16(\text{positive number}) < 0$. This means $f(x)$ is decreasing for $x > 0$. 4. **Identify the extremum:** Since $f'(x)$ changes from positive to negative at $x=0$, there is a local maximum at $x=0$. The value of the function at this point is $f(0) = -2(0)^8 = 0$. Furthermore, for any $x \neq 0$, $x^8 > 0$, so $-2x^8 < 0$. Since $f(0) = 0$ and $f(x) < 0$ for all other $x$, $(0,0)$ is a global maximum. The correct statement is: The function $f(x)$ has a global maximum at $x=0$.

Every differentiable function is continuous

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for x=1, LHL=1 and RHL=5