12×6=(8×6)+(_×6)=answer this question

Answer:

The correct option is 3.

Step-by-step explanation:

If a straight line passes through some of the points, none of the points, or all of the points of a scatter plot and best represents the data, then it is called a best fit line.

Best fit line minimizes the squared distance of each point. Some points of the scatter plot lie above and below the best fit line.

When determining slope, data points do not usually fall within a straight line, so you need to draw a best fit line, with as many points above and below the line as possible.

Therefore the correct option is 3.

The woman should have paid \$64 for the regular price of the merchandise.

To find out what the woman should have paid for the regular price, we need to reverse engineer the discount. Here's the step-by-step calculation:

Calculate the Discount Amount: Since the sale is 1/4 off, the discount is 1/4 of the regular price. So, the discount is:

[tex]\[ \frac{1}{4} \times \text{Regular Price} \][/tex]

[tex]\[ \text{Discount} = \frac{1}{4} \times \text{Regular Price} \][/tex]

Subtract the Discount from the Regular Price to Get the Sale Price: The sale price is the regular price minus the discount:

[tex]\[ \text{Sale Price} = \text{Regular Price} - \text{Discount} \][/tex]

Calculate the Sale Price: We know the woman bought $48 worth of merchandise at the sale price, so:

[tex]\[ \text{Sale Price} = \$48 \][/tex]

Find the Regular Price: Now we can set up an equation using the information we have:

[tex]\[ \$48 = \text{Regular Price} - \text{Discount} \][/tex]

Substituting the discount calculated earlier:

[tex]\[ \$48 = \text{Regular Price} - \left(\frac{1}{4} \times \text{Regular Price}\right) \][/tex]

Solve for the Regular Price:

[tex]\[ \$48 = \left(\frac{4}{4}\right) \times \text{Regular Price} - \left(\frac{1}{4}\right) \times \text{Regular Price} \][/tex]

[tex]\[ \$48 = \left(\frac{3}{4}\right) \times \text{Regular Price} \][/tex]

To isolate the regular price, multiply both sides by [tex]\( \frac{4}{3} \)[/tex]:

[tex]\[ \text{Regular Price} = \$48 \times \frac{4}{3} \][/tex]

[tex]\[ \text{Regular Price} = \$64 \][/tex]

So, the woman should have paid $64 for the regular price of the merchandise.

Complete Question:

The dress store is having a sale where all merchandise is 1/4 off a woman buys $48 of merchandise at the sale price what should she have paid for the regular price

Answer:

18.5 meters

Step-by-step explanation:

As the maximum height ona projectile motion that starts from the ground and finishes at the ground is simmetric, the maximum height of the projectile will be found exactly at half of the distance tranveled horizontally, so the exact half of the total distance traveled horizontally is 18.5 meters, since distance is 37 meters and divided by 2 is 18.5.

The ball will travel 18.5 meters to reach its maximum height.

663/3 = 21

21 is the middle number,

21+1 = 22

21-1 = 20

20 + 21 + 22 = 63

Our three consecutive integers can be represented as

X → first integer

X + 1 → second integer

X + 2 → third integer

If the sum of our three consecutive integers is 63, our equation will read...

X + X + 1 + X + 2 = 63

Now, we can simplify on the left. Once we simplify, our equation will now read

3x + 3 = 63

      -3    -3  ← subtract 3 on both sides

  3x = 60 ← divide both sides by 3

   X = 20

Our three consecutive integers can be represented as

X → first integer = 20

X + 1 → second integer = 21

X + 2 → third integer = 22

Therefore, the three consecutive integers are 20, 21, and 22.

To solve the given system of equations using substitution, we'll start by solving the first equation for x in terms of y. We'll substitute this expression for x in the other two equations to find the values of y and z.

To solve the given system of equations using substitution, we'll start by solving the first equation for x in terms of y: x = -2y + 10. We'll substitute this expression for x in the other two equations to find the values of y and z. Substituting -2y + 10 for x in the second equation, we have -6y = 6(-2y + 10) - 6.

Simplifying this equation, we get y = -2. Substituting y = -2 in the first equation, we find x = -2*(-2) + 10 = 14. Finally, substituting x = 14 and y = -2 in the third equation, we find z = -3*14 + 10*(-2) = -52.

.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 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.

Answer:29.15475947 (not rounded)

Step-by-step explanation: