All Answers 355859 Answers

Use the two labelled points to calculate the slope  =   (-5-4) / ( -1 -5) =  9/6 

In   y = mx + b form this becomes     y = 9/6 x + b     sub in any point on the line to find b = - 3.5 

so the line is      y = 9/6 x - 3 1/2     re-arrange 

                          - 9/6 x + y = - 3 1/5 

                                                              divide thru by - 3 1/2 

                             3/7 x  - 2/7 y  = 1      so a = 3/7 

Vertex at 3,1    and a point on the parabola is  0, 7 

Vertex form :

   f(x) =  d ( x-3)^2 + 1       sub on the point 0,7   to find  d = 2/3

so the equation is now 

    f(x) = 2/3(x-3)^2 + 1      which expands to    2/3 x^2 -4 + 7      so    a = 2/3       b = - 4    c = 7    which when multiplied together = -18 2/3

Analyzing the Problem

We have a quadrilateral BCED with extended sides meeting at point A. We're given several side lengths and need to find DE.

Solution Approach

To find DE, we'll need to break down the quadrilateral into triangles and use similar triangles to establish proportions.

Breaking Down the Quadrilateral

Let's divide quadrilateral BCED into two triangles: Triangle ABC and Triangle ADE.

Using Similar Triangles

We can see that triangle ABC and triangle ADE share angle A. Additionally, angle ABC is congruent to angle ADE (alternate interior angles since BD is parallel to AE). Therefore, triangles ABC and ADE are similar.

Setting up Proportions

Since the triangles are similar, their corresponding sides are proportional. We can set up the following proportion:

AB / BD = AC / DE

Substituting Values and Solving

We know AB = 3, BD = 18, and AC = 7. Substituting these values into the proportion, we get:

3 / 18 = 7 / DE

Cross-multiplying, we get:

3 * DE = 18 * 7

DE = (18 * 7) / 3

DE = 42

Therefore, DE = 42.

First, let's use the Law of Cosines. It states that \( cos PMQ = - cos PMR \)

Thus, we can write the equations

\(PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)\)

Plugging in the values we already know from the problem, we get

\(5^2 = 5.5^2 + PM^2 + 2(5.5 * PM)cos(PMR) \\ 8^2 = 5.5^2 + PM^2 - 2(5,5 * PM) cos (PMR)       \)

Now, add these two equations. We get

\(5^2 + 8^2  = 2 * 5.5^2 + 2PM^2 \\ 89 = 60.5 + 2PM^2 \\ 28.5 / 2 = PM^2 \\ 14.25 = PM^2 \)

Thus, we have \(\sqrt{14.25}  = PM ≈  3.77\)

The answer is 3.77. 

Thanks! :)

Proving Equidistance of Intersection Points on an Archimedean Spiral

Understanding the Problem

We're considering the Archimedean spiral defined by the polar equation r = θ for θ > 0. The task is to prove that consecutive intersection points of this spiral with any ray originating from the origin are equidistant.

Solution

1. Parametric Representation of the Ray:

Let the ray be defined by the angle φ.

Any point on this ray can be represented in polar coordinates as (r, φ), where r is the distance from the origin.

2. Intersection Points:

To find the intersection points of the spiral and the ray, we equate their polar equations:

θ = r

Substituting r with r = ρcos(φ - θ) (polar coordinate conversion to Cartesian), we get:

θ = ρcos(φ - θ)

3. Finding Consecutive Intersection Points:

Let θ₁ and θ₂ be the angles corresponding to two consecutive intersection points.

Then, we have:

θ₁ = ρcos(φ - θ₁)

θ₂ = ρcos(φ - θ₂)

4. Calculating the Distance Between Intersection Points:

The distance between two points in polar coordinates is given by:

d = sqrt(r₁² + r₂² - 2r₁r₂cos(θ₂ - θ₁))

Substituting r₁ = θ₁ and r₂ = θ₂, we get:

d = sqrt(θ₁² + θ₂² - 2θ₁θ₂cos(θ₂ - θ₁))

5. Simplifying the Distance:

Using the trigonometric identity cos(A - B) = cosAcosB + sinAsinB, we can simplify the expression for d:

d = sqrt((θ₂ - θ₁)²)

d = |θ₂ - θ₁|

6. Conclusion:

Since θ₂ and θ₁ represent consecutive intersection points, |θ₂ - θ₁| is a constant (equal to 2π for a full rotation).

Therefore, d = 2π is also a constant.

The Distance

The distance between consecutive intersection points of the Archimedean spiral with any ray from the origin is 2π.

This proves that the consecutive intersection points are equidistant.

Let  S  represent the weight of one sack of sugar.

Let  F  represent the weight of one sack of flour.

Two sacks of suagr together with three sacks of lour weight no more than 40 pounds:

    2S + 5F <= 60

The weight of a sack of flour is no more than 5 poiunds more than the weight of two sacks of sugar:

    F <= 2S + 8

Since both of the inequalities have the same sense (point in the same direction), we can substitute:

Substituting:  2S + 3(2S + 5) <= 40

                        2S + 6S + 15 <= 40

                                 8S + 15 <= 40

                                         8S <= 25

                                           S <= 3.125

Since  F <= 2S + 5   --->   F <= 2(3.125) + 5   --->   F <= 11.25

So, each sack of flour cannot weigh more than 11.25 pounds.

First, let's simplify and combine all like terms. 

After doing that, we get \(3x^2-kx+33=0\)

Now, in order for the equation to have no real solutions, the descriminant must be less than 0. 

The descriminant is \(b^2-4ac\), so we have the equation \(k^2 - 4(33)(3) < 0\)

Since we are trying to find the largest, we set the two to equal each other and get

\(k^2 - 396 = 0\\ k^2 = 396\\ k \approx 19.89974\)

Rounding down, the largest k can be as an integer is 19. 

So our answer is 19. 

Thanks! :)

First, let's combine all like terms and put this in the form of a quadratic equation. We have

\(3x^{2}+4x+10\)

Now, the descriminant is in the form of \(b^2-4ac\) for qudratics in the form \(ax^2+bx+c\)

Thus, we have

 \(4^2-4(3)(10)\\ = 16 - 120\\ -104\)

So the answer is -104. 

Thanks! :)

Good answer by EP here

I do not think this is a geometric series. 

We have 

\(0.0000324/0.036=0.0009\)

\(0.036/0.4=0.09\)

mmmm...🤔🤔🤔

I think this is correct....

Thanks! :)