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To find $\frac{dy}{dx}$ for the implicit function $x^3 + y^3 = 5$ , we use implicit differentiation with respect to $x$ .

Differentiate each term with respect to x:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(5)$

Apply the power rule for $x^3$ and the chain rule for $y^3$ :

$3x^2 + 3y^2 \frac{dy}{dx} = 0$

Rearrange the equation to solve for $\frac{dy}{dx}$ :

$3y^2 \frac{dy}{dx} = -3x^2$

Divide both sides by 3y^2:

$\frac{dy}{dx} = \frac{-3x^2}{3y^2}$

Simplify the expression:

$\frac{dy}{dx} = -\frac{x^2}{y^2}$

Using the property of logarithms $\log_b (MN) = \log_b M + \log_b N$ :

$\log_2 8 + \log_2 4 = \log_2 (8 \times 4) = \log_2 32$.

Alternatively, we can evaluate each logarithm separately:

$\log_2 8 = 3$ (since $2^3 = 8$ )

$\log_2 4 = 2$ (since $2^2 = 4$ )

So, $3 + 2 = 5$.

For a quadratic function $y = ax^2 + bx + c$ , the sign of the coefficient of $x^2$ ($a$) determines the direction of opening of the parabola.

If $a > 0$ , the parabola opens upwards (U-shaped), indicating a minimum point.

If $a < 0$ , the parabola opens downwards (inverted U-shaped), indicating a maximum point

Explanation:

1. Median: The data set has 8 values (an even number). The median is the average of the two middle values, which are the 4th and 5th terms (x and 40).

⇒ $\text{Median} = \frac{x + 40}{2}$

⇒ Given, Median = 35.

$35 = \frac{x + 40}{2}$

$70 = x + 40$

$x = 70 - 40 = 30$

2. Mean: The mean is the sum of all values divided by the count.

⇒ $\text{Sum} = 20 + 25 + 30 + x + 40 + y + 55 + 60$

⇒ Substitute x=30:

$\text{Sum} = 20 + 25 + 30 + 30 + 40 + y + 55 + 60 = 260 + y$

⇒ Given, Mean = 40.

⇒ $40 = \frac{260 + y}{8}$

⇒ $320 = 260 + y$

⇒ $y = 320 - 260 = 60$

Therefore, x = 30 and y = 60.

Explanation:

⇒ Using the product rule of logarithms: $\log_b X + \log_b Y = \log_b (X \times Y)$

$\log_a 4 + \log_a 16 = \log_a (4 \times 16) = \log_a 64$

⇒ So the equation becomes:

$\log_a 64 = 3$

⇒ By definition of logarithm, if $\log_a X = Y$ , then $a^Y = X$ .

⇒ So, $a^3 = 64.$

⇒ To find a, take the cube root of 64:

⇒ $a = \sqrt[3]{64}$

⇒ Since $4 \times 4 \times 4 = 64$ ,

⇒ a = 4

Let's examine each statement:

A. Opposite sides are equal in length: This is a property of all parallelograms, and thus also true for a rhombus (which is a type of parallelogram). (True for both)

B. Diagonals bisect each other: This is a property of all parallelograms, and thus also true for a rhombus. (True for both)

C. All four sides are equal in length: This is the defining property of a rhombus. A general parallelogram only requires opposite sides to be equal. (True for rhombus, not necessarily for general parallelogram)

D. Opposite angles are equal: This is a property of all parallelograms, and thus also true for a rhombus. (True for both)

Therefore, the statement that is always true for a rhombus but not necessarily for a general parallelogram is that all four sides are equal in length.

The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Given the points (-2, 5) and (6, -3):

$x_1 = -2, y_1 = 5$

$x_2 = 6, y_2 = -3$

Calculate the x-coordinate of the midpoint:

$x_{\text{mid}} = \frac{-2 + 6}{2} = \frac{4}{2} = 2$

Calculate the y-coordinate of the midpoint:

$y_{\text{mid}} = \frac{5 + (-3)}{2} = \frac{2}{2} = 1$

The coordinates of the midpoint are (2, 1).

Substitute a=3 and b=2 into the given definition of the binary operation:

$3 * 2 = 3^2 + 2^2 - (3)(2)$

$3 * 2 = 9 + 4 - 6$

$3 * 2 = 13 - 6$

$3 * 2 = 7$.

According to the Factor Theorem, if a polynomial$P(x)$ is divisible by ,$(x-a)$ then $(x-a)$ is a factor of ,$P(x)$ which means that when$x=a$ is substituted into the polynomial, the result is zero, i.e., $P(a)=0$.

Explanation:

1. Evaluate the terms with exponents:

$3^{-1} = \frac{1}{3}$

$2^2 = 4$

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

2. Substitute these values into the expression:

$\frac{\frac{1}{3} + 4}{\frac{1}{4}}$

3. Simplify the numerator:

$\frac{1}{3} + 4 = \frac{1}{3} + \frac{12}{3} = \frac{13}{3}$

4. Perform the division:

$\frac{\frac{13}{3}}{\frac{1}{4}} = \frac{13}{3} \times \frac{4}{1} = \frac{52}{3}$

Use the Sophie Germain identity:$X^4 + 4Y^4 = (X^2 + 2Y^2 + 2XY)(X^2 + 2Y^2 - 2XY)$ .

Here, we have $a^4 + 64$.

We can write $64 = 4 \times 16 = 4 \times (2)^4.$

So, $X=a$ and $Y=2$ .

Substitute into the identity:

$a^4 + 4(2)^4 = (a^2 + 2(2)^2 + 2a(2))(a^2 + 2(2)^2 - 2a(2))$

$a^4 + 64 = (a^2 + 8 + 4a)(a^2 + 8 - 4a)$

Rearrange the terms:

$a^4 + 64 = (a^2 + 4a + 8)(a^2 - 4a + 8)$

For a $2 \times 2 matrix A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is given by $ad - bc$ .

For matrix $A = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ ,

Determinant of $A = (2)(3) - (1)(4)$

$= 6 - 4$

$= 2$ .

First, convert the fractions to decimals or common fractions:

$\frac{2}{0.05} = \frac{2}{\frac{5}{100}} = \frac{200}{5} = 40$

==> $\frac{1}{0.015} = \frac{1}{\frac{15}{1000}} = \frac{1000}{15} = \frac{200}{3}$

Now substitute these values into the expression:

$\left|40 - \frac{200}{3}\right| = \left|\frac{120}{3} - \frac{200}{3}\right| = \left|-\frac{80}{3}\right|$

The absolute value is:

$\left|-\frac{80}{3}\right| = \frac{80}{3}$

Convert to decimal and round to 2 decimal places:

$\frac{80}{3} \approx 26.666... \approx 26.67$

Explanation: The mode of a data set is the value that appears most frequently.

In the given data set:

⇒ 15 appears 2 times

⇒ 12 appears 3 times

⇒ 18 appears 2 times

⇒ 10 appears 1 time

The value 12 appears more often than any other value. Therefore, the mode is 12.

According to the Remainder Theorem:

When $P(x)$ is divided by $(x-2)$ , the remainder is $P(2) = 7$ .

When $P(x)$ is divided by $(x+1)$ , the remainder is $P(-1) = 1$ .

Let the remainder when $P(x)$ is divided by $(x-2)(x+1)$ be $Ax + B$ , since the divisor is quadratic.

So, $P(x) = Q(x)(x-2)(x+1) + Ax + B$ .

Using the values from the Remainder Theorem:

For $x=2$ : $P(2) = A(2) + B \Rightarrow 2A + B = 7$ (Equation 1)

For $x=-1$ : $P(-1) = A(-1) + B \Rightarrow -A + B = 1$ (Equation 2)

Subtract Equation 2 from Equation 1:

$(2A + B) - (-A + B) = 7 - 1$

$2A + B + A - B = 6$

$3A = 6$

$A = 2$ .

Substitute $A=2$ into Equation 2:

$-2 + B = 1$

$B = 1 + 2 = 3$ .

So, the remainder is $Ax + B = 2x + 3$ .

Given equations:

1. 2x + 3y = 12

2. x - y = 1

From equation (2), express x in terms of y:

$x = 1 + y \quad \text{(Equation 3)}$

Substitute Equation 3 into Equation 1:

2(1 + y) + 3y = 12

Expand and simplify:

2 + 2y + 3y = 12

2 + 5y = 12

Subtract 2 from both sides:

5y = 12 - 2

5y = 10

$y = \frac{10}{5}$

y = 2

Substitute y = 2 back into Equation 3 to find x:

x = 1 + 2

x = 3

The solution is x=3 and y=2.

1. Find the reference angle:

First, find the acute angle (reference angle, let's call it $\beta$ ) for which $\cos \beta = \frac{\sqrt{3}}{2}$ .

$\cos \beta = \frac{\sqrt{3}}{2} \implies \beta = 30^\circ$

2. Determine the quadrants:

Since $\cos \alpha$ is negative $(-\frac{\sqrt{3}}{2})$ , $\alpha$ must lie in the quadrants where cosine is negative. These are the second (Q2) and third (Q3) quadrants.

3. Calculate $\alpha$ in Q2:

In the second quadrant, $\alpha$ $= 180^\circ - \beta$ .

$\alpha_1 = 180^\circ - 30^\circ = 150^\circ$

4. Calculate $\alpha$ in Q3:**

In the third quadrant, $\alpha = 180^\circ + \beta$ .

$\alpha_2 = 180^\circ + 30^\circ = 210^\circ$

Therefore, the values of $\alpha$ for which $\cos \alpha = -\frac{\sqrt{3}}{2}$ in the given range are $150^\circ$ and $210^\circ.$

We need to find $\frac{dV}{dt}$ . We are given $\frac{dr}{dt} = 0.5 \text{ cm/s}$.

First, differentiate V with respect to r:

$\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi (3r^2) = 4\pi r^2$ .

Using the chain rule, $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$ .

Substitute the values: $r=4 \text{ cm}$ and $\frac{dr}{dt} = 0.5 \text{ cm/s}$ .

$\frac{dV}{dt} = (4\pi (4)^2) \times 0.5$

$\frac{dV}{dt} = (4\pi \times 16) \times 0.5$

$\frac{dV}{dt} = 64\pi \times 0.5$

$\frac{dV}{dt} = 32\pi \text{ cm}^3/\text{s}$ .

The formula for the sum to infinity of a G.P. is $S_\infty = \frac{a}{1-r}$ , where $a$ is the first term and $r$ is the common ratio, provided $|r| < 1$ .

Given $a = 16$ and $r = \frac{1}{2}$ . Since $|\frac{1}{2}| < 1$ , the sum to infinity exists.

$S_\infty = \frac{16}{1 - \frac{1}{2}}$

$S_\infty = \frac{16}{\frac{1}{2}}$

$S_\infty = 16 \times 2$

$S_\infty = 32$

If we substitute $x=2$ directly, we get $\frac{0}{0}$ , which is an indeterminate form.

Factorize the numerator using the difference of two squares formula ($a^2 - b^2 = (a-b)(a+b)$):

$x^2 - 4 = (x-2)(x+2)$ .

So, $\lim_{x \to 2} \frac{(x-2)(x+2)}{x - 2}$ .

Cancel out the common term $(x-2)$ , assuming $x \neq 2$ :

$\lim_{x \to 2} (x+2)$.

Now, substitute $x=2$ :

$2 + 2 = 4$