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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:

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

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.

Explanation: Follow the order of operations (parentheses/brackets, exponents/orders, multiplication and division (from left to right), and addition and subtraction (from left to right)).

1. Multiplication inside the first parenthesis:

$(\frac{1}{2} * 2) = 1$

2. Division inside the second parenthesis:

$(6 \div \frac{2}{3}) =(6 \times \frac{3}{2}) = \frac{18}{2} = 9$

3. Substitute back into the expression:$5 - 1 + 9$

4. Perform addition and subtraction from left to right: $4 + 9 = 13$

Total hours worked = 40 hours.

Overtime hours = h hours.

Standard work hours = Total hours - Overtime hours = 40 - h hours.

Earning from standard work = Standard hours \\times standard rate

$\text{Standard Earning} = (40 - h) \times p = (40 - h)p$

Earning from overtime = Overtime hours x overtime rate (1.5p)

==> $\text{Overtime Earning} = h \times (1.5p) = 1.5hp$

Total weekly earning = Standard Earning + Overtime Earning

==> $\text{Total Earning} = (40 - h)p + 1.5hp$

Expand the expression:

$\text{Total Earning} = 40p - hp + 1.5hp$

Combine like terms:

$\text{Total Earning} = 40p + 0.5hp$

Factor out p:

$\text{Total Earning} = p(40 + 0.5h)$

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:

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

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.

A bar chart is commonly used for representing categorical data, where each bar represents a distinct category and its height (or length) indicates the frequency or count of items in that category.

1. Histograms are used for continuous data.
2. Line graphs are used to show trends over time or continuous variables.
3. Pie charts are used to show parts of a whole, often for categorical data, but a bar chart is generally more versatile for comparison across multiple categories.

1. Calculate the volume of the large container:**

Radius of container $R = diameter / 2 = 12cm / 2 = 6cm.$

Height of container H = 20cm.

Volume of container $V\_{cont} = \\pi R^2 H$

==> $V_{\text{cont}} = \pi (6\text{ cm})^2 (20\text{ cm}) = \pi (36\text{ cm}^2) (20\text{ cm}) = 720\pi \text{ cm}^3$

2. Calculate the volume of one small jar:

Radius of jar $r = diameter / 2 = 3cm / 2 = 1.5cm.$

Height of jar $h = 10cm.$

Volume of jar $V\_{jar} = \\pi r^2 h$

==> $V_{\text{jar}} = \pi (1.5\text{ cm})^2 (10\text{ cm}) = \pi (2.25\text{ cm}^2) (10\text{ cm}) = 22.5\pi \text{ cm}^3$

3. Calculate how many small jars can be filled:

Number of jars = $V\_{cont} / V\_{jar}$

==> $\text{Number of jars} = \frac{720\pi \text{ cm}^3}{22.5\pi \text{ cm}^3} = \frac{720}{22.5} = 32$

4. Calculate total revenue from selling the honey:

Revenue = Number of jars times price per jar

==> $\text{Revenue} = 32 \times \text{₦}20.00 = \text{₦}640.00$

5. Calculate the profit (amount made):

Profit = Revenue - Cost Price

==> $\text{Profit} = \text{₦}640.00 - \text{₦}100.00 = \text{₦}540.00$

The permutation formula is $nP_r = \frac{n\!}{(n-r)\!}$ .

The given ratio can be written as:

$\frac{12P_{k+1}}{11P_{k+1}} = \frac{12}{5}$

Expand the permutation terms:

$\frac{\frac{12!}{(12 - (k+1))!}}{\frac{11!}{(11 - (k+1))!}} = \frac{12}{5}$

Simplify the denominators:

$\frac{\frac{12!}{(11 - k)!}}{\frac{11!}{ (10 - k)!}} = \frac{12}{5}$

Rewrite the division as multiplication by the reciprocal:

$\frac{12!}{(11 - k)!} \times \frac{(10 - k)!}{11!} = \frac{12}{5}$

Use the property $n\! = n \times (n-1)\!$ :

$12\! = 12 \times 11\!$

$(11 - k)\! = (11 - k) \times (10 - k)\!$

Substitute these into the equation:

$\frac{12 \times 11!}{(11 - k) \times (10 - k)!} \times \frac{(10 - k)!}{11!} = \frac{12}{5}$

Cancel out $11\! and (10 - k)\!$ :

$\frac{12}{11 - k} = \frac{12}{5}$

Since the numerators are equal (12 on both sides), the denominators must also be equal:

11 - k = 5

Solve for k:

k = 11 - 5

k = 6

Explanation:

1. Total possible outcomes: When two dice are tossed, the total number of possible outcomes is 6 \times 6 = 36

2. Possible sums (from 2 to 12):

3. Identify prime numbers among these sums: The prime numbers between 2 and 12 are 2, 3, 5, 7, 11.

4. List outcomes that sum to these prime numbers:

⇒ Sum = 2: (1,1) - 1 outcome

⇒ Sum = 3: (1,2), (2,1) - 2 outcomes

⇒ Sum = 5: (1,4), (4,1), (2,3), (3,2) - 4 outcomes

⇒ Sum = 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) - 6 outcomes

⇒ Sum = 11: (5,6), (6,5) - 2 outcomes

5. Total favorable outcomes: 1 + 2 + 4 + 6 + 2 = 15 outcomes.

6. Calculate the probability:

$\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{15}{36}$

7. Simplify the fraction: Divide both numerator and denominator by their greatest common divisor, 3.

$\frac{15 \div 3}{36 \div 3} = \frac{5}{12}$

Explanation:

1. Calculate the Loss:

$\text{Loss} = \text{Cost Price} - \text{Selling Price}$

$\text{Loss} = \text{₦}80,000 - \text{₦}64,000 = \text{₦}16,000$

2. Calculate Percentage Loss:

$\text{Percentage Loss} = \frac{\text{Loss}}{\text{Cost Price}} \times 100\%$

$\text{Percentage Loss} = \frac{\text{₦}16,000}{\text{₦}80,000} \times 100\%$

$\text{Percentage Loss} = \frac{16}{80} \times 100\%$

$\text{Percentage Loss} = \frac{1}{5} \times 100\%$

$\text{Percentage Loss} = 20\%$

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 $\frac{dr}{dt} = 0.5 \text{ cm/s}$ .

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

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

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

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

1. Convert volume from litres to $cm^3$ :

We know that 1 litre = 1000 $cm^3$ .

So, $385 litres = 385 \times 1000 \text{ cm}^3 = 385000 \text{ cm}^3$ .

$V = 385000 \text{ cm}^3$

2. Calculate the radius of the tank:

Diameter = 70 cm.

$Radius (r) = Diameter / 2 = 70 cm / 2 = 35 cm.$

$r = 35 \text{ cm}$

3. Use the formula for the volume of a cylinder to find height (h):

$Volume (V) = \pi r^2 h$

$V = \pi r^2 h$

Rearrange to solve for h:

$h = \frac{V}{\pi r^2}$

4. Substitute the values and calculate h:

Use $\pi = \frac{22}{7}.$

$h = \frac{385000}{\frac{22}{7} \times (35)^2}$

$h = \frac{385000}{\frac{22}{7} \times 1225}$

$h = \frac{385000}{22 \times \frac{1225}{7}}$

$h = \frac{385000}{22 \times 175}$

$h = \frac{385000}{3850}$

$h = 100 \text{ cm}$

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

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

$3 * 2 = 9 + 4 - 6$

$3 * 2 = 13 - 6$

$3 * 2 = 7$.

Explanation:

⇒ To differentiate $y = 3\cos(5x)$ , we use the chain rule.

⇒ Let u = 5x. Then $\frac{du}{dx} = 5$ .

⇒ The function becomes $y = 3\cos(u)$ .

⇒ Differentiate y with respect to u:

$\frac{dy}{du} = 3(-\sin(u)) = -3\sin(u)$ .

⇒ By the chain rule,

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ .

$\frac{dy}{dx} = (-3\sin(u)) \times 5 = -15\sin(u)$

⇒ Substitute u = 5x back: $\frac{dy}{dx} = -15\sin(5x)$ .

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 $x=2$ :

$2 + 2 = 4$

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).

The relationship is $P = k_1 Q + k_2 Q^2$ .

Using $P=10, Q=1$ :

$10 = k_1(1) + k_2(1)^2 \Rightarrow 10 = k_1 + k_2$ (Equation 1)

Using $P=28, Q=2$ :

$28 = k_1(2) + k_2(2)^2 \Rightarrow 28 = 2k_1 + 4k_2$ (Equation 2)

From Equation 1, $k_1 = 10 - k_2$ . Substitute into Equation 2:

$28 = 2(10 - k_2) + 4k_2$

$28 = 20 - 2k_2 + 4k_2$

$28 - 20 = 2k_2$

$8 = 2k_2 \Rightarrow k_2 = 4$ .

Substitute $k_2=4$ into $k_1 = 10 - k_2$ :

$k_1 = 10 - 4 = 6$ .

So, the relationship is $P = 6Q + 4Q^2$ .

⇒ Since DE is parallel to BC, by the property of similar triangles, triangle ADE is similar to triangle ABC.

⇒ Therefore, the ratio of corresponding sides is equal:

$\frac{AD}{AB} = \frac{AE}{AC}$

⇒ From the diagram, AD = 2.

⇒ The full length of side AB is AB = AD + DB = 2 + 6 = 8.

⇒ We are given AC = 10.

⇒ Substitute these values into the ratio:

$\frac{2}{8} = \frac{AE}{10}$

⇒ Simplify the fraction:

$\frac{1}{4} = \frac{AE}{10}$

⇒ Cross-multiply:

$4 \times AE = 1 \times 10$

$4AE = 10$

$AE = \frac{10}{4} = \frac{5}{2} = 2.5$

1. Find the total number of ratio parts:

The ratio of mangoes for Ade : Ben : Charity is 3:4:5.

Total parts in the ratio = 3 + 4 + 5 = 12 parts.

$\text{Total parts} = 3 + 4 + 5 = 12$

2. Determine the value of one ratio part:

Ben's share is 4 parts, and Ben received 20 mangoes.

So, 4 parts = 20 mangoes.

$1 part = \frac{20}{4} mangoes = 5 mangoes.$

$1 \text{ part} = \frac{20 \text{ mangoes}}{4 \text{ parts}} = 5 \text{ mangoes/part}$

3. Calculate the total number of mangoes shared out:

Multiply the total ratio parts by the value of one part:

$\text{Total mangoes} = \text{Total parts} \times \text{Value of 1 part}$

$\text{Total mangoes} = 12 \times 5 = 60 \text{ mangoes}$