Determine whether quantities vary directly or inversely and find the constant of variation. A teacher grades 25 students essays in 4 hours. Assuming he grades at the same speed, how long will it take him to grade 35 essays?

Answer:

ANSWER IS

Step-by-step explanation:

the answer to your question is

 Unit walking rate in miles per hour = 5 2/3 / 2 2/3 = 17/3 / 8/3 = 17/3 x 3/8 = 17/8 = 2 1/8 miles per hour.

Unit walking rate in hours per mile = 1/ 17/8 = 8/17 hours per mile

Answer:

Rachel enjoys exercising outdoors. Today she walked 5 2/3 miles or 5.67 miles in 2 2/3 hours or 2.67 hours.

Rachel’s unit walking rate in miles per hour is = [tex]\frac{5.67}{2.67}[/tex] = 2.125 miles per hour or 2 1/8 miles per hour.

Rachel’s unit walking rate in hours per mile = [tex]\frac{2.67}{5.67}[/tex] = 0.47 hours per mile.

The B. distance of a javelin throw is an example of a continuous random variable.

A continuous random variable is a variable that can take any value within a specified range, often associated with measurements and quantities. Unlike discrete variables, it can assume an infinite number of values, typically represented by intervals on the real number line.

Out of the given options, the distance of a javelin throw is an example of a continuous random variable. A continuous random variable is one that can take on any value within a given range. In the case of a javelin throw, the distance can be any real number value within a certain range, such as 0 to infinity. This is in contrast to the other options, which are discrete random variables that can only take on specific whole number values.

Length is [tex]\boldsymbol{4.4}[/tex] feets and width is equal to [tex]\boldsymbol{8.8}[/tex] feets.

The area of a two-dimensional region, form, or planar lamina in the plane is the quantity that expresses its extent.

A quadrilateral having four right angles is known as a rectangle.

Let [tex]\boldsymbol{w}[/tex] feets denotes width of a rectangle.

Length of a rectangle [tex]=\boldsymbol{w-4}[/tex] feets

Area of a rectangle [tex]=\boldsymbol{42}[/tex] square feet

Length [tex]\times[/tex] Width [tex]=42[/tex]

      [tex]w(w-4)=42[/tex]

[tex]w^2-4w-42=0[/tex]

[tex]w=\frac{4\pm \sqrt{16+168}}{2}[/tex]

   [tex]=2\pm \sqrt{46}[/tex]

As dimension can not be negative, [tex]2-\sqrt{46}[/tex] is rejected.

So,

[tex]w=2+ \sqrt{46}[/tex]

   [tex]=\boldsymbol{8.8}[/tex] feets

Length [tex]=8.8-4[/tex]

            [tex]=\boldsymbol{4.4}[/tex] feets

So, length is [tex]4.4[/tex] feets and width is equal to [tex]8.8[/tex] feets.

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Group terms that contain the same variable, and move the constant to the opposite side of the equation

Rewrite as perfect squares

Answer:

Option 4.

Step-by-step explanation:

To find the value of 'b' that gives the maximum value of 86 for the function f(x) = -x² + bx - 14, substitute the x-coordinate of the vertex (-b/2a) into the function, equate it with 86 and solve for 'b'.

To find the values of b for which the quadratic function f(x) = -x² + bx - 14; Maximum value: 86 holds true, we need to utilize the properties of quadratic functions. In a quadratic function in the form f(x) = ax² + bx + c, the maximum value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by -b/2a.

Given that our quadratic is a maximum (opens downwards since a=-1), the y-coordinate of the vertex, which is our maximum value, is 86. Substituting these into our function, we get f(-b/2a) = 86, replacing a=-1, this reduces to -b²/4 -14 = 86. Simplifying this equation will give you the value of b.

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The perimeters of two rhombi with equal areas are sometimes equal, but not always.

Let's consider the formula for the area of a rhombus, which is given by [tex]\( A = \frac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \)[/tex] are the lengths of the diagonals. For the perimeter P, we have [tex]\( P = 4 \times s \)[/tex], where s is the length of one side of the rhombus.

For two rhombi to have equal areas, the product of their diagonals must be the same. That is, if we have two rhombi with diagonals [tex]\( (d_1, d_2) \) and \( (d_1', d_2') \), then \( d_1 \times d_2 = d_1' \times d_2' \)[/tex].

However, the perimeter depends only on the length of one side, s, and not on the diagonals. Therefore, two rhombi with the same area can have different side lengths and thus different perimeters.

Let's consider an example:

Rhombus 1:

[tex]- Diagonals \( d_1 = 8 \) units and \( d_2 = 4 \) units[/tex]

[tex]- Area \( A = \frac{1}{2} \times 8 \times 4 = 16 \) square units[/tex]

[tex]- Side length \( s \), using the Pythagorean theorem (since the diagonals bisect each other at right angles), is \( s = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \) units[/tex]

[tex]- Perimeter \( P = 4 \times s = 4 \times \sqrt{20} \) units[/tex]

Rhombus 2:

[tex]- Diagonals \( d_1' = 10 \) units and \( d_2' = 3.2 \) units[/tex]

[tex]- Area \( A' = \frac{1}{2} \times 10 \times 3.2 = 16 \) square units (equal to the area of Rhombus 1)[/tex]

[tex]- Side length \( s' \), using the Pythagorean theorem, is \( s' = \sqrt{1.6^2 + 2.56^2} = \sqrt{2.56 + 6.5536} = \sqrt{9.1136} \) units[/tex]

[tex]- Perimeter \( P' = 4 \times s' = 4 \times \sqrt{9.1136} \) units[/tex]

Comparing the perimeters:

[tex]- \( P = 4 \times \sqrt{20} \approx 4 \times 4.472 = 17.888 \) units[/tex]

[tex]- \( P' = 4 \times \sqrt{9.1136} \approx 4 \times 3.0188 = 12.0752 \) units[/tex]

Clearly, the perimeters are not equal, even though the areas are the same.

However, it is possible for two rhombi with equal areas to have equal perimeters if their side lengths are the same. For instance, if Rhombus 1 and Rhombus 2 both had side lengths of s = s' = 4 units, then their perimeters would both be P = P' = [tex]4 \times 4 = 16 \)[/tex] units, regardless of the lengths of their diagonals, as long as the diagonals satisfy the area condition [tex]\( d_1 \times d_2 = d_1' \times d_2' \)[/tex].

In conclusion, the perimeters of two rhombi with equal areas can sometimes be equal, specifically when the side lengths are the same, but they are not always equal, as demonstrated by the example above.

a² + b² = c²

c² - a² = b²

17² - 7² = x²   ↓

289 - 49 = √240

√240 = 15.49193338482967 (rounded 15.49 or 15.5)

Hope this helps,

♥A.W.E.S.W.A.N.♥

Answer:

The other factors are (x-1) and (x+2).

The algebraic expression is D = 50t where D is the distance covered in miles and t is the time taken in hours.

Distance is defined as the product of speed and time.

Arithmetic operations can also be specified by adding, subtracting, dividing, and multiplying built-in functions.

The operator that performs the arithmetic operations is called arithmetic operator.

* Multiplication operation: Multiplies values on either side of the operator

For example 4*2 = 8

/ Division operation: Divides left-hand operand by right-hand operand

For example 4/2 = 2

Given that the distance a car travels in t hours at a rate of 50 miles per hour

Let the distance covered in miles = D

And t is the time taken in hours

Since distance is defined as the multiplication of speed and time.

So an algebraic expression for the verbal description as

⇒ D = 50t

Hence, the algebraic expression is D = 50t where D is the distance covered in miles and t is the time taken in hours.

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Answer:

I think it's b, please correct me if I'm wrong.

Step-by-step explanation:

I'm taking the test rn so I hope its right.

Answer:

its b i just did it

Step-by-step explanation:

Answer:  The required values are

[tex]f^{-1}(x)=x-4~~~~~~~~\textup{and}~~~~~~~~~f^{-1}(4)=0.[/tex]

Step-by-step explanation:  We are given the following function f(x) :

[tex]f(x)=x+4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We are to find the values of [tex]f^{-1}(x)[/tex] and [tex]f^{-1}(4).[/tex]

Let us consider that

[tex]y=f(x)~~~~~~~~\Rightarrow x=f^{-1}(y).[/tex]

So, from equation (i), we get

[tex]f(x)=x+4\\\\\Rightarrow y=f^{-1}(y)+4\\\\\Rightarrow f^{-1}(y)=y-4\\\\\Rightarrow f^{-1}(x)=x-4.[/tex]

Substituting x = 4 in the above equation, we get

[tex]f^{-1}(4)=4-4=0\\\\\Rightarrow f^{-1}(4)=0.[/tex]

Thus, the required values are

[tex]f^{-1}(x)=x-4~~~~~~~~\textup{and}~~~~~~~~~f^{-1}(4)=0.[/tex]

Answer:

Rewrite the inequality so there is a single rational expression on one side, and 0 on the other.

Combine under a common denominator.

Test points in the critical regions.

Construct the solution.

Step-by-step explanation:

Answer Above