Which transformation were applied to the graph of the parent function y=tan(x) to produce the function graphed below?

A figure is composed of a rectangle and a right triangle. What is the area of the figure?

strojnjashka [21]

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

$A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h$

We substitute the values to obtain;

$Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm$

$Area=\frac{1}{2}(21.6cm)\times 6cm$

$Area=64.8cm^2$

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

$C=2\pi r$

where $r=3.5ft$ is the radius of the circular table cloth.

We substitute this value and $\pi=3.14$ into the formula to get;

$C=2(3.14)(3.5)ft$

$C=21.98ft$

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

$C=\pi r$

where $r=12.5cm$ is the radius of the semicircle.

$C=3.14\times 12.5cm$

$C=39.25cm$

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

$Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh$

$Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24$

$Area=345.3125+300$

$Area=645.3125cm^2$

$Area\approx645cm^2$

QUESTION 6

The area of the parallelogram is given by the formula;

$A=bh$

From the diagram, $b=y\:in.,h=3in.$ .

Given, A=23.7 inches squared.

We substitute the values to obtain;

$23.7=3y$

$y=\frac{23.7}{3}$

$y=7.9in.$

QUESTION 7

The area of a rectangle is given by the formula;

$Area=l\times b$

The area of the bigger rectangle $=9\times15=135in^2$

The area of the smaller rectangle $=8\times 13=104in^2$

The area of the shaded region $=135-104=31in^2$

5 0

3 years ago

7) -6x + 6y=6 -6x + 3y=-12

Gre4nikov [31]

Answer:

(5,6)

Step-by-step explanation:

-6x+6y=6

-6x + 3y =-12

Multiply the first equation by -1

6x-6y=-6

Add this to the second equation

6x-6y=-6

-6x + 3y =-12

---------------------

-3y = -18

Divide each side by -3

-3y/-3 = -18/-3

y =6

Now we need to find x

6x - 6y = -6

6x -6(6) = -6

6x -36 = -6

Add 36 to each side

6x-36+36 = -6+36

6x = 30

Divide each side by 6

6x/6 = 30/6

x =5

3 0

4 years ago

Miguel is reading a 338-page book. In the first 40 minutes he spends reading, he reads 52 pages. If he continues reading at the

viktelen [127]

Answer: 260 minutes.

Step-by-step explanation:

This represents the ratio between the minutes he reads and the pages he reads: 40 : 52

The book Miguel is reading is 338 pages long. Miguel also reads 52 pages in 40 minutes. You can divide 338 by 52, which is equal to 6.5.

If it takes Miguel 40 minutes to read 52 pages, then we can multiply that by 6.5. 40 x 6.5 = 260.

In conclusion, it will take Miguel 260 minutes to read the entire book, or 4 hours and 20 minutes.

4 0

3 years ago

Read 2 more answers

In a random sample of 63 women at a company, the mean salary is $48,902 with a standard deviation of $5270. in a rando sample of

sdas [7]

So u would make at least 101,000 per year

4 0

3 years ago

A path 5m wide is to be built along the border and inside a square garden of side 90m. Find the cost of cementing the path at th

Leya [2.2K]

Answer:

Total cost of cementing the path = Rs. 17,000.

Step-by-step explanation:

Let the square garden be PQRS

Let the region inside the garden (PQRS) be KLMN.

Given the following data;

Length of sides of PQRS = 90m

Width of path = 5m

Cost = Rs. 10 per m²

Area of PQRS = 90 * 90

Area of PQRS = 8100m²

To find the area of KLMN;

KL = KN = 90 - (5 + 5)

KL = KN = 90 - 10

KL = KN = 80m

Area of KLMN = KL * KN

Substituting into the equation, we have;

Area of KLMN = 80 * 80

Area of KLMN = 6400 m²

Area of path = Area of PQRS - Area of KLMN

Area of path = 8100 - 6400

Area of path = 1700 m²

Total cost of cementing the path = Area of path * Cost

Total cost of cementing the path = 1700 * 10

Total cost of cementing the path = Rs. 17,000

8 0

3 years ago