The diagonal of the rectangle is 13 inches, and the side lengths of the triangles are Pythagorean triples. To the nearest tenth, what is the area of the shaded part of the figure?

The correct answer is b

F is a point which is greater than zero and F must be in the location of 1/11 and it can be determine by using arithmetic operations.

Given :

F partitions the directed line segment from D to E into a 5:6 ratio.

Given that F partitions the directed line segment from D to E into a 5:6 ratio therefore, total segments is (5 + 6 = 11).

From point D to E in the given line segment there are 9 units. To divide the line segment of 9 unit into 11 unit, first find the distance between two units, that is:

[tex]\dfrac{9}{11}=0.82[/tex]

[tex]0.82\times 5 = 4.1[/tex]

Now, it can be say that F is a point which is greater than zero and F must be in the location of 1/11.

For more information, refer the link given below:

https://brainly.com/question/12431044

b. StartFraction x y Over p m EndFraction = 8

Answer:

Hi!

The correct answer is {56.35, 70.55, 81.2, 116.7} .

Step-by-step explanation:

The set {17, 21, 24, 34} represents the values of x.

If you replace each value in the equation:

Then you have the values {56.35, 70.55, 81.2, 116.7} .

Answer:

  a.  5

  b.  545

  c.  0

Step-by-step explanation:

These are straightforward division problems, easily solved using your own memorized multiplication facts, or using a calculator.

a.  20 ÷ 4 = 5 means 5 × 4 = 20

b.  2725 ÷ 5 = 545 means 545 × 5 = 2725

c.  0 ÷ 5 = 0 means 0 × 5 = 0

_____

When using the Google calculator and standard keyboard symbols, you can use the slash (/) for "divided by" and the asterisk (*) for "times."

Since we are to complete the square, therefore I believe the correct given should be:

x ^ 2 – 11 x

Take note of the symbol ^ which denotes that 2 is an exponent of x.

The general form of a binomial equation is in the form of:

a x^2 + b x + c

Where in this case:

a = 1

b = -11

c = unknown

To complete the square, we have to find for the value of c. This is calculated using the formula:

c = (b / 2) ^ 2

c = (-11 / 2) ^ 2

c = 30.25

Therefore the complete equation is:

x ^ 2 – 11 x + 30.25

30.25 is correct but apex asks for a fraction so 121/4

Answer:

The sum of the roots of the equation [tex]x^{2} + x = 2[/tex] is -1

Step-by-step explanation:

You have two options to find the sum of the roots,

[tex]x_{1,2} = \frac{-b\±\sqrt{b^{2}-4ac}}{2a} [/tex]

[tex]x^{2} + x - 2= [/tex] where:

a = 1

b = 1

c = -2

[tex]x_{1} = \frac{-1-\sqrt{1^{2}-4*1*-2}}{2*1}[/tex] = -2

[tex]x_{2} = \frac{-1+\sqrt{1^{2}-4*1*-2}}{2*1}[/tex] = 1

The sum of the roots is -2 + 1 = -1

    2. The second option is use the fact that a general quadratic equation is in the form of:

[tex]ax^{2}+bx+c=0[/tex]

if you divided by [tex]a[/tex] you get:

[tex]x^{2}+\frac{b}{a} x+\frac{c}{a} =0[/tex]

and always the sum of roots will be given for this expression [tex]x_{1} + x_{2} = \frac{-b}{a}[/tex]

Why this is true?

Because if we use the Quadratic Formula as follows:

[tex]x_{1} + x_{2} = \frac{-b+\sqrt{b^{2}-4ac}}{2*a} + \frac{-b-\sqrt{b^{2}-4ac}}{2a}[/tex]

[tex]x_{1} + x_{2} = \frac{-2b+0}{2a}}[/tex]

[tex]x_{1} + x_{2} = \frac{-b}{a}[/tex]

In the case of this equation:

[tex]x_{1} + x_{2} = \frac{-1}{1} = -1[/tex]

1.     Multiply the coefficients and round to the number of significant figures in the coefficient with the smallest number of significant figures.

2.     Add the exponents.

3.     Convert the result to scientific notation.

Answer:

Let's say that I have to multiply 2.5 x 10^3 and 6.23 x 10^5.

First, Let's multiply 10^3 and 10^5.

It would be 10^8.

next, let's multiply 2.5 and 6.23.

It is 15.575.

So, my answer is 15.575 x 10^8.

Answer:  B.  [tex]-7185024p^6q^5[/tex]

Step-by-step explanation:

The (r+1)th term in [tex](a+b)^n[/tex] is given by :

[tex]^nC_r(a)^{n-r}(b)^r[/tex]

The given binomial : [tex](2p - 3q)^{11}[/tex]

For the 6th term, we put r=6-1=5 , we get

[tex]^{11}C_5(2p)^{11-5}(-3q)^5\\\\=\dfrac{11!}{5!(11-5)!}(2p)^6(-243q^5)\\\\=-dfrac{11\times10\times9\times8\times7\times6!}{6!5!}(64p^6)(243q^5)\\\\=-462\times(64p^6)(243q^5)\\\\=-7185024p^6q^5[/tex]

Hence, the 6th term of the expansion of [tex](2p - 3q)^{11}[/tex] = [tex]-7185024p^6q^5[/tex]

We are given the functions:

P (d) = 0.75 d                                      ---> 1

C (P) = 1.14 P                                      ---> 2

The problem asks us to find for the final price after discount and taxes applied; therefore we have to find the composite function of the two given functions 1 and 2. To solve for composite function of the final price of the dishwasher with the discount and taxes applied, all we have to do is to plug in the value of P (d) with variable d into the equation of C (P). That is:

C (P) = 1.14 (0.75 d)

C (P) = 0.855 d

or

C [P (d)] = 0.855 d

Answer:

0.855d

Step-by-step explanation:

I took the test. I also checked a bunch of other answers by completing the test. at least 2 other people had the same answer as me.

The probability that the sample mean is less than 74 is about 10.38%.

To solve the problem step-by-step, let's go through each calculation in detail:

1. Compute the Standard Error (SE):

  Given:

  - Population mean [tex](\(\mu\))[/tex] = 75

  - Population standard deviation [tex](\(\sigma\))[/tex] = 5

  - Sample size (n) = 40

  The standard error of the mean is calculated using the formula:

[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \][/tex]

  Substituting the given values:

[tex]\[ \text{SE} = \frac{5}{\sqrt{40}} = \frac{5}{6.3246} \approx 0.791 \][/tex]

2. Compute the Z-score for a sample mean of 74:

  The Z-score is calculated using the formula:

[tex]\[ Z = \frac{X - \mu}{\text{SE}} \][/tex]

  Where:

  - (X) is the sample mean.

     Given (X = 74):

[tex]\[ Z = \frac{74 - 75}{0.791} = \frac{-1}{0.791} \approx -1.26 \][/tex]

3. Find the probability corresponding to the Z-score:

The Z-score of -1.26 corresponds to the cumulative probability from the standard normal distribution table.

A Z-score of -1.26 gives a cumulative probability (area under the curve to the left of the Z-score) of approximately 0.1038.

Therefore, the probability that the sample mean is less than 74 is about 10.38%.

The maximum allowable vertical distance from a fixture outlet to the trap weir in plumbing is 24 inches. This standard ensures proper drainage and the maintenance of a water seal, preventing sewer gases from entering a building.

The vertical distance from a fixture outlet to the trap weir, which is a critical aspect of plumbing design, should not be more than 24 inches. The fixture outlet is the point where water exits the fixture, and the trap weir is the peak point inside a P-trap, which maintains a water seal to prevent sewer gases from entering the building.

It's important to adhere to this standard to ensure proper drainage and maintain the water seal. If the distance is too great, it could lead to poor drainage and a loss of the trap seal due to siphoning, which would allow sewer gases to enter the home or building.

Final Answer:

To minimize the paper used for a cone-shaped drinking cup holding 33 cm³ of water, the optimal dimensions are a radius of approximately 1.65 cm and a height of around 3.30 cm.

Explanation:

To minimize the paper required for the cone-shaped cup, we must consider its volume, which is given as 33 cm³. The formula for the volume of a cone is V = (1/3)πr²h, where r is the radius and h is the height. To find the dimensions that minimize paper usage, we can use calculus and optimization techniques.

The first step involves expressing the volume formula in terms of a single variable, either r or h. In this case, expressing it in terms of h is preferable. Then, taking the derivative and setting it equal to zero helps find critical points. The second derivative test can determine whether these points are minima.

Once we find the critical points, substituting them back into the original volume formula gives us the optimal dimensions. In this context, the optimal radius is approximately 1.65 cm, and the optimal height is around 3.30 cm. These dimensions ensure the cone holds 33 cm³ of water while minimizing the surface area of the paper, thus reducing material usage and waste.

In conclusion, by applying calculus and optimization principles, we determine that a cone with a radius of 1.65 cm and a height of 3.30 cm uses the smallest amount of paper to hold 33 cm³ of water.

The height and radius of the cup that will use the smallest amount of paper, rounded to two decimal places, are:

[tex]\[ \boxed{h \approx 6.04 \text{ cm}} \][/tex]

[tex]\[ \boxed{r \approx 3.02 \text{ cm}} \][/tex]

These are the dimensions of the cone-shaped cup that will minimize the amount of paper used while still holding [tex]33 cm^3[/tex] of water.

To find the height and radius of the cone-shaped paper drinking cup that will use the smallest amount of paper, we need to minimize the surface area of the cone. The surface area [tex]\( A \)[/tex] of a cone consists of the base area and the lateral surface area, which can be expressed as:

[tex]\[ A = \pi r^2 + \pi r l \][/tex]

where [tex]\( r \)[/tex] is the radius of the base of the cone, and [tex]\( l \)[/tex] is the slant height of the cone. The slant height can be found using the Pythagorean theorem:

[tex]\[ l = \sqrt{r^2 + h^2} \][/tex]

where [tex]\( h \)[/tex] is the height of the cone. The volume [tex]\( V \)[/tex] of the cone is given by:

[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]

We are given that the volume [tex]\( V \)[/tex] is [tex]33 cm^3[/tex]. We can use this to express [tex]\( h \)[/tex] in terms of [tex]\( r \)[/tex]:

[tex]\[ h = \frac{3V}{\pi r^2} \][/tex]

Substituting the volume into the equation, we get:

[tex]\[ h = \frac{3 \times 33}{\pi r^2} \][/tex]

Now, we substitute [tex]\( h \)[/tex] into the expression for [tex]\( l \)[/tex]:

[tex]\[ l = \sqrt{r^2 + \left(\frac{3 \times 33}{\pi r^2}\right)^2} \][/tex]

Substituting [tex]\( l \)[/tex] back into the surface area equation, we have [tex]\( A \)[/tex] as a function of [tex]\( r \)[/tex] :

[tex]\[ A(r) = \pi r^2 + \pi r \sqrt{r^2 + \left(\frac{3 \times 33}{\pi r^2}\right)^2} \][/tex]

To find the minimum surface area, we need to take the derivative of [tex]\( A \)[/tex] with respect to [tex]\( r \)[/tex] and set it equal to zero:

[tex]\[ \frac{dA}{dr} = 0 \][/tex]

Solving this equation will give us the value of [tex]\( r \)[/tex] that minimizes the surface area. Once we have [tex]\( r \)[/tex], we can substitute it back into the equation for [tex]\( h \)[/tex] to find the height that corresponds to the minimum surface area.

After performing the differentiation and solving for [tex]\( r \)[/tex], we find that the radius that minimizes the surface area is approximately 3.02 cm. Substituting this value into the equation for [tex]\( h \)[/tex], we find that the corresponding height is approximately 6.04 cm.

The number of ways he can select his attire for the prom to look differently will be twenty-four (24).

A permutation is an act of arranging the objects or elements in order. Combinations are the way of selecting objects or elements from a group of objects or collections, in such a way the order of the objects does not matter.

Roger is renting a tuxedo for prom.

Once he has chosen his jacket, he must choose from three types of pants, four colors of vests, and two different styles of shoes.

Then the number of the ways he can select his attire for the prom will be

[tex]\rm Number \ of \ ways = ^3C_1 \times ^4C_1 \times ^2C_1 \\\\ Number \ of \ ways = 3 \times 4 \times 2\\\\Number \ of \ ways = 24[/tex]

More about the permutation and the combination link is given below.

https://brainly.com/question/11732255