Your parents have a credit card with a balance of $3,287.90 at an interest rate of 14.5% APR. They pay $1,200.00 each month on the due date until the card is paid off. How many months does it take to pay off the card, and what is the total amount paid including interest?

Be sure to include in your response:
• the answer to the original question
• the mathematical steps for solving the problem demonstrating mathematical reasoning

Answer:

Option 3

(3xy³ + 5x²y⁴)  (3xy³ - 5x²y⁴)

Step-by-step explanation:

Factorize polynomials:

Use exponent law:

                   [tex]\boxed{\bf a^{m*n}=(a^m)^n} \ & \\\\\boxed{\bf a^m * b^m = (a*b)^m}[/tex]

9x²y⁶ = 3²* x² * y³*² = 3² * x² * (y³)² = (3xy³)²

25x⁴y⁸ = 5² * x²*² * y⁴*² = 5² * (x²)² * (y⁴)² = (5x²y⁴)²

Now use the identity:  a² - b² = (a +b) (a -b)

Here, a = 3xy³ & b = 5x²y⁴

9x²y⁶ - 25x⁴y⁸ = 3²x²(y³)² - 5²(x²)² (y⁴)²

                       = (3xy³)² - (5x²y⁴)²

                      = (3xy³ + 5x²y⁴)  (3xy³ - 5x²y⁴)

When solving a linear system of equations, you are looking for the points of intersection between the equations

From the question, we have the following parameters that can be used in our computation:

Solving a system of linear equations

Also, we have the following from the options

The general rile is that

Slope = rate of change

x and y intercepts = when y and x equals 0

points of intersection = solution to the system

Hence, you are looking for the points of intersection between the equations

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Question

When solving a linear system of equations, you are looking for which of the following?

Slope

y-intercept

x-intercept

Points of intersection

The value of integral is 3.

Let's evaluate the integral over the positive half of the interval:

∫[0 to π] (cos(x) / √(4 + 3sin(x))) dx

Let u = 4 + 3sin(x), then du = 3cos(x) dx.

Substituting these expressions into the integral, we have:

∫[0 to π] (cos(x) / sqrt(4 + 3sin(x))) dx = ∫[0 to π] (1 / (3√u)) du

Using the power rule of integration, the integral becomes:

∫[0 to π] (1 / (3√u)) du = (2/3) . 2√u ∣[0 to π]

Evaluating the definite integral at the limits of integration:

(2/3)2√u ∣[0 to π] = (2/3) 2(√(4 + 3sin(π)) - √(4 + 3sin(0)))

(2/3) x 2(√(4) - √(4)) = (2/3) x 2(2 - 2) = (2/3) x 2(0) = 0

So, the value of integral is

= 3-0

= 3

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Answer:

[tex]3-\displaystyle \int^{\pi}_{-\pi} \dfrac{\cos x}{\sqrt{4+3 \sin x}}\; \text{d}x\approx 0.806\; \sf (3\;d.p.)[/tex]

Step-by-step explanation:

First, compute the indefinite integral:

[tex]\displaystyle \int \dfrac{\cos x}{\sqrt{4+3\sin x}}\; \text{d}x[/tex]

To evaluate the indefinite integral, use the method of substitution.

[tex]\textsf{Let} \;\;u = 4 + 3 \sin x[/tex]

Find du/dx and rewrite it so that dx is on its own:

[tex]\dfrac{\text{d}u}{\text{d}x}=3 \cos x \implies \text{d}x=\dfrac{1}{3 \cos x}\; \text{d}u[/tex]

Rewrite the original integral in terms of u and du, and evaluate:

[tex]\begin{aligned}\displaystyle \int \dfrac{\cos x}{\sqrt{4+3\sin x}}\; \text{d}x&=\int \dfrac{\cos x}{\sqrt{u}}\cdot \dfrac{1}{3 \cos x}\; \text{d}u\\\\&=\int \dfrac{1}{3\sqrt{u}}\; \text{d}u\\\\&=\int\dfrac{1}{3}u^{-\frac{1}{2}}\; \text{d}u\\\\&=\dfrac{1}{-\frac{1}{2}+1} \cdot \dfrac{1}{3}u^{-\frac{1}{2}+1}+C\\\\&=\dfrac{2}{3}\sqrt{u}+C\end{aligned}[/tex]

Substitute back u = 4 + 3 sin x:

                            [tex]= \dfrac{2}{3}\sqrt{4+3\sin x}+C[/tex]

Therefore:

[tex]\displaystyle \int \dfrac{\cos x}{\sqrt{4+3\sin x}}\; \text{d}x= \dfrac{2}{3}\sqrt{4+3\sin x}+C[/tex]

To evaluate the definite integral, we must first determine any intervals within the given interval -π ≤ x ≤ π where the curve lies below the x-axis. This is because when we integrate a function that lies below the x-axis, it will give a negative area value.

Find the x-intercepts by setting the function to zero and solving for x in the given interval -π ≤ x ≤ π.

[tex]\begin{aligned}\dfrac{\cos x}{\sqrt{4+3\sin x}}&=0\\\\\cos x&=0\\\\x&=\arccos0\\\\\implies x&=-\dfrac{\pi }{2}, \dfrac{\pi }{2}\end{aligned}[/tex]

Therefore, the curve of the function is:

So to calculate the total area, we need to calculate the positive and negative areas separately and then add them together, remembering that if you integrate a function to find an area that lies below the x-axis, it will give a negative value.

Integrate the function between -π and -π/2.

As the area is below the x-axis, we need to negate the integral so that the resulting area is positive:

[tex]\begin{aligned}A_1=-\displaystyle \int^{-\frac{\pi}{2}}_{-\pi} \dfrac{\cos x}{\sqrt{4+3\sin x}}\; \text{d}x&=- \left[\dfrac{2}{3}\sqrt{4+3\sin x}\right]^{-\frac{\pi}{2}}_{-\pi}\\\\&=-\dfrac{2}{3}\sqrt{4+3 \sin \left(-\frac{\pi}{2}\right)}+\dfrac{2}{3}\sqrt{4+3 \sin \left(-\pi\right)}\\\\&=-\dfrac{2}{3}\sqrt{4+3 (-1)}+\dfrac{2}{3}\sqrt{4+3 (0)}\\\\&=-\dfrac{2}{3}+\dfrac{4}{3}\\\\&=\dfrac{2}{3}\end{aligned}[/tex]

Integrate the function between -π/2 and π/2:

[tex]\begin{aligned}A_2=\displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \dfrac{\cos x}{\sqrt{4+3\sin x}}\; \text{d}x&= \left[\dfrac{2}{3}\sqrt{4+3\sin x}\right]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\\\\&=\dfrac{2}{3}\sqrt{4+3 \sin \left(\frac{\pi}{2}\right)}-\dfrac{2}{3}\sqrt{4+3 \sin \left(-\frac{\pi}{2}\right)}\\\\&=\dfrac{2}{3}\sqrt{4+3 (1)}-\dfrac{2}{3}\sqrt{4+3 (-1)}\\\\&=\dfrac{2\sqrt{7}}{3}-\dfrac{2}{3}\\\\&=\dfrac{2\sqrt{7}-2}{3}\end{aligned}[/tex]

Integrate the function between π/2 and π.

As the area is below the x-axis, we need to negate the integral so that the resulting area is positive:

[tex]\begin{aligned}A_3=-\displaystyle \int^{\pi}_{\frac{\pi}{2}} \dfrac{\cos x}{\sqrt{4+3\sin x}}\; \text{d}x&= -\left[\dfrac{2}{3}\sqrt{4+3\sin x}\right]^{\pi}_{\frac{\pi}{2}}\\\\&=-\dfrac{2}{3}\sqrt{4+3 \sin \left(\pi\right)}+\dfrac{2}{3}\sqrt{4+3 \sin \left(\frac{\pi}{2}\right)}\\\\&=-\dfrac{2}{3}\sqrt{4+3 (0)}+\dfrac{2}{3}\sqrt{4+3 (1)}\\\\&=-\dfrac{4}{3}+\dfrac{2\sqrt{7}}{3}\\\\&=\dfrac{2\sqrt{7}-4}{3}\end{aligned}[/tex]

To evaluate the definite integral, sum A₁, A₂ and A₃:

[tex]\begin{aligned}\displaystyle \int^{\pi}_{-\pi} \dfrac{\cos x}{\sqrt{4+3\sin x}}\; \text{d}x&=\dfrac{2}{3}+\dfrac{2\sqrt{7}-2}{3}+\dfrac{2\sqrt{7}-4}{3}\\\\&=\dfrac{2+2\sqrt{7}-2+2\sqrt{7}-4}{3}\\\\&=\dfrac{4\sqrt{7}-4}{3}\\\\ &\approx2.194\; \sf (3\;d.p.)\end{aligned}[/tex]

Now we have evaluated the definite integral, we can subtract it from 3 to evaluate the given expression:

[tex]\begin{aligned}3-\displaystyle \int^{\pi}_{-\pi} \dfrac{\cos x}{\sqrt{4+3 \sin x}}\; \text{d}x&=3-\dfrac{4\sqrt{7}-4}{3}\\\\&=\dfrac{9}{3}-\dfrac{4\sqrt{7}-4}{3}\\\\&=\dfrac{9-(4\sqrt{7}-4)}{3}\\\\&=\dfrac{13-4\sqrt{7}}{3}\\\\&\approx 0.806\; \sf (3\;d.p.)\end{aligned}[/tex]

Therefore, the given expression cannot be zero.

The area of the dining room table top is: 864 sq. in

The formula for the area of a triangle is:

Area = ¹/₂ * base * height

Formula for the area of a rectangle is:

Area = Length * Width

Area of trapezium = ¹/₂(sum of parallel sides) * height

Thus, if 1cm = 6 inches

Then: 9cm = 54 inches

3 cm = 18 inches

4 cm = 24 inches

Thus:

Area of trapezium = ¹/₂(54 + 18) * 24

= 864 sq. in

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The quantity of goods and services that sellers are willing and able to sell is known as the quantity supplied.

Quantity supplied is defined as the amount of goods and services that a supplier is able to produce and sell at a given market price.

The quantity supplied differs from the actual amount of supply as price changes influence how much supply producers actually put on the market.

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