Graph the image of the given triangle after the transformation with the rule (x, y)→(x−1, y+2) . Select the "Polygon" button from the tool bar to plot your triangle. You may use the "Move" button to move the triangle after it is created. PolygonMove UndoRedoReset

Final answer:

To find the expression with a quotient greater than 1, we convert the mixed numbers into improper fractions and perform the division for each expression. The expression that gives a quotient greater than 1 is 5 3/4 divided by 7 3/8.

Explanation:

To find the expression that has a quotient greater than 1, we need to divide the given fractions. Let's examine each expression:

4 2/3 divided by 5 1/4: To divide mixed numbers, we convert them into improper fractions and then perform the division. In this case, the quotient is less than 1.

3 1/8 divided by 2 2/5: Similar to the first expression, we convert the mixed numbers into improper fractions and divide. The quotient is also less than 1.

1 6/7 divided by 2 1/3: Again, convert the mixed numbers into improper fractions and perform the division. The quotient is less than 1.

5 3/4 divided by 7 3/8: Convert the mixed numbers into improper fractions and divide. The quotient is greater than 1, so this is the expression we're looking for.

Therefore, the expression with a quotient greater than 1 is 5 3/4 divided by 7 3/8.

The numerator of the equivalent fraction with denominator 12 for 1/2 is 6. This comes from multiplying both the numerator and denominator of 1/2 by the same number (6).

To find the numerator of the equivalent fraction with denominator 12 for 1/2, we simply multiply both parts of the fraction by the same number such that the denominator becomes 12. We know that 2 * 6 equals 12, so we multiply the numerator 1 by 6 as well, which gives us the numerator 6.

Therefore, the fraction 1/2 is equivalent to 6/12.

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.002 is 1/10 of 0.02. To find this, we multiply .002 by 10. This is seen as moving the decimal point to the right once because we are multiplying by a power of ten.

The student asked: ".002 is 1/10 of what". To find the answer to this question, we can set up a simple equation where we multiply .002 by 10 because we know that if .002 is 1/10 of a number, then multiplying it by 10 will give us that whole number.

So the equation is:

Therefore, .002 is 1/10 of 0.02.

This can also be understood by recognizing that when dividing by powers of 10, you move the decimal to the left by the number of zeros in the power of ten. So, in the reverse process, when we want to find a number that is ten times larger (i.e., 1/10 of the original), we move the decimal one place to the right.

Answer:  5

Step-by-step explanation:

In the given picture we have a table representing two columns as x and y.

We know that the slope of the function is given by :-

[tex]k=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

Using points (0,4) and (2,14) from the table , we get the slope of the function will be :-

[tex]k=\dfrac{14-4}{2-0}\\\\\\=\dfrac{10}{2}=5[/tex]

Therefore, the slope of the given function in the table = 5

The value of the expression [tex]16 - x^{2}[/tex] is 16 cos^2θ

Here,

The expression is [tex]16 - x^{2}[/tex]

We have to find the value of expression after substitute 4 sinθ for x.

What is substitution method?

Substitution method is algebraic method to solve the linear equations.



Now,

The expression is [tex]16 - x^{2}[/tex]

By putting x = 4 sinθ in above equation, we get;

[tex]16 - x^{2}[/tex]  = 16 - ( 4 sinθ)^2

            = 16 - 16 sin^2θ

            = 16 ( 1 - sin^2θ )

            = 16 cos^2θ

Since,  1 - sin^2θ = cos^2θ

Hence, The value of the expression [tex]16 - x^{2}[/tex] is 16 cos^2θ

Learn more about the substitution method visit:

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To simplify the expression 16 - x² after substituting 4 sin θ for x, you end up with 16 cos² θ, using the Pythagorean trigonometric identity.

To simplify the expression 16 - x² after substituting 4 sin θ for x, where 0° < θ < 90°, we follow these steps:

The substitution gives us 16 - (4 sin θ)².

To further simplify:

48/12 = 4 feet

24/2 = 2 feet

 4 *2 = 8 square feet

Answer:

                     8 feet squared

The area of the region is [tex]\(\frac{64}{3}\)[/tex] square units.

To find the area of the region bounded by the parabola \(y = x^2\), the tangent line at the point (2, 4), and the x-axis, we'll first determine the points of intersection.

The tangent line at (2, 4) has the same slope as the derivative of the parabola at x = 2. The derivative of [tex]\(y = x^2\)[/tex] is [tex]\(2x\)[/tex], so at x = 2, the slope is [tex]\(2 \times 2 = 4\)[/tex]. Thus, the equation of the tangent line is [tex]\(y - 4 = 4(x - 2)\)[/tex].

Now, let's find the points of intersection between the parabola and the tangent line:

[tex]\[x^2 = 4(x - 2) + 4\][/tex]

Solving for x, we get [tex]\(x^2 - 4x = 0\)[/tex], which factors to [tex]\(x(x - 4) = 0\)[/tex]. So, [tex]\(x = 0\)[/tex] and [tex]\(x = 4\)[/tex].

Now, integrate the absolute difference of the functions from 0 to 4 to find the area:

[tex]\[A = \int_0^4 |x^2 - (4x - 4)| \,dx.\][/tex]

Evaluate the integral:

[tex]\[A = \int_0^4 (x^2 - 4x + 4) \,dx = \frac{1}{3}x^3 - 2x^2 + 4x \Big|_0^4 = \frac{64}{3}.\][/tex]

Therefore, the area of the region is [tex]\(\frac{64}{3}\)[/tex] square units.

The answer is: [tex]\frac{8}{3}[/tex].

The area of the region bounded by the parabola [tex]\( y = x^2 \)[/tex], the tangent line to this parabola at the point (2, 4), and the x-axis is given by the integral of the difference between the parabola and the tangent line from [tex]\( x = 0 \) to \( x = 2 \)[/tex].

First, we need to find the equation of the tangent line to the parabola at the point (2, 4). The slope of the tangent line is the derivative of the parabola's equation at \( x = 2 \). The derivative of [tex]\( y = x^2 \)[/tex] is [tex]\( y' = 2x \).[/tex] At[tex]\( x = 2 \)[/tex], the slope is [tex]\( 2 \cdot 2 = 4 \)[/tex].

Using the point-slope form of a line,[tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( (x_1, y_1) \)[/tex] is a point on the line, we get the equation of the tangent line as follows:

[tex]\[ y - 4 = 4(x - 2) \][/tex]

[tex]\[ y = 4x - 8 + 4 \][/tex]

[tex]\[ y = 4x - 4 \][/tex]

Now, we will find the area under the parabola and above the tangent line from [tex]\( x = 0 \) to \( x = 2 \)[/tex]. The integral of [tex]\( x^2 \)[/tex] from 0 to 2 is:

[tex]\[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \][/tex]

The integral of the tangent line \( y = 4x - 4 \) from 0 to 2 is:

[tex]\[ \int_{0}^{2} (4x - 4) \, dx = \left[ 2x^2 - 4x \right]_{0}^{2} = (2 \cdot 2^2 - 4 \cdot 2) - (2 \cdot 0^2 - 4 \cdot 0) = (8 - 8) - (0 - 0) = 0 \][/tex]

The area between the parabola and the tangent line is the difference between these two integrals:

[tex]\[ \text{Area} = \int_{0}^{2} x^2 \, dx - \int_{0}^{2} (4x - 4) \, dx \][/tex]

[tex]\[ \text{Area} = \frac{8}{3} - 0 \][/tex]

[tex]\[ \text{Area} = \frac{8}{3} \][/tex]

Therefore, the area of the region bounded by the parabola [tex]\( y = x^2 \)[/tex], the tangent line at the point (2, 4), and the x-axis is [tex]\( \boxed{\frac{8}{3}} \)[/tex] square units.