What is the best estimate for the product 289 and 7? Make it clear to understand : )

Help please !!! At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating

Radda [10]

Answer:

n = 12$ x = 8$ y = 10$ where n, x, and y are number of tickets

12 n + 8 x + 10 y = 3920 and n + x + y = 420

12n + 8 (x + y) + 2 y = 3920

12 n + 8 (5 n) + 2 y = 3920 since 5 (x + y) = n

52 n + 2 y = 3920 or y = 1960 - 26 n

Also, n + x + y = 420 or n + 5 n = 420 since x + y = 5 n

n = 70 so 70 of the $12 were sold

And since y = 1960 - 26 n we have y = 140 tickets

Now 12 * 70 + 8 x + 140 * 10 = 3920

This gives x = 210 tickets

Check: 210 + 140 + 70 = 420 tickets

Also, 12 * 70 + 210 * 8 + 140 * 10 = 3920

7 0

3 years ago

The latest mystery novel cost $24 the table shows the sales of that this novel by a bookstore what was the dollar amount of the

Ket [755]

Answer:

$3768

Step-by-step explanation:

The amount of sales will be the product of the selling price of each book and the number of books sold.

sales = book price × number of books

sales = $24 × 157 . . . . . . . . books sold on Saturday

sales = $3768 . . . . . . . . . . . sales on Saturday

4 0

3 years ago

SUPER URGENT: Complete the general form of the equation of a sinusoidal function having an amplitude of 6, a period of 2pi/3, an

mel-nik [20]

Answer:

y = 6·sin(3·(x - 1)) + c

Step-by-step explanation:

The general form of an equation for a sinusoidal function is presented ad follows;

y = a·sin(b·(x - h) + c

Where;

a = The amplitude of the equation

T = The period = 2·π/b

h = The phase shift

c = The vertical shift

From the question, we have;

a = 6,

2·π/3 = 2·π/b

∴ b = 3

h = 1

We get;

y = 6·sin(3·(x - 1)) + c.

4 0

3 years ago

Find all the missing sides or angles in each right triangles

astra-53 [7]

In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:

Parallel Postulate: Given: line l and a point P not on l. There is exactly one line through P that is parallel to l.

In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:

Triangle △XYZ is cut by AB¯¯¯¯¯¯¯¯ where A and B are midpoints of sides XZ¯¯¯¯¯¯¯¯ and YZ¯¯¯¯¯¯¯ respectively. AB¯¯¯¯¯¯¯¯ is called a midsegment of △XYZ. Note that △XYZ has other midsegments in addition to AB¯¯¯¯¯¯¯¯. Can you see where they are in the figure above?

If we construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and construct CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ respectively, we have the following figure and see that segments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ are midsegments of △XYZ.

In this lesson we will investigate properties of these segments and solve a variety of problems.

Properties of midsegments within triangles

We start with a theorem that we will use to solve problems that involve midsegments of triangles.

Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

parallel to the third side. half as long as the third side.

Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.

Consider the following triangle △XYZ. Construct the midpoint A of side XZ¯¯¯¯¯¯¯¯.

By the Parallel Postulate, there is exactly one line though A that is parallel to side XY¯¯¯¯¯¯¯¯. Let’s say that it intersects side YZ¯¯¯¯¯¯¯ at point B. We will show that B must be the midpoint of XY¯¯¯¯¯¯¯¯ and then we can conclude that AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is parallel to XY¯¯¯¯¯¯¯¯.

We must show that the line through A and parallel to side XY¯¯¯¯¯¯¯¯ will intersect side YZ¯¯¯¯¯¯¯ at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point B is the midpoint of side YZ¯¯¯¯¯¯¯.

Since XA¯¯¯¯¯¯¯¯≅AZ¯¯¯¯¯¯¯, we have BZ¯¯¯¯¯¯¯≅BY¯¯¯¯¯¯¯¯. Hence, by the definition of midpoint, point B is the midpoint of side YZ¯¯¯¯¯¯¯. AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is also parallel to XY¯¯¯¯¯¯¯¯.

Proof of 2. We must show that AB=12XY.

In △XYZ, construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and midsegments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ as follows:

First note that CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ by part one of the theorem. Since CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ and AB¯¯¯¯¯¯¯¯∥XY¯¯¯¯¯¯¯¯, then ∠XAC≅∠BCA and ∠CAB≅∠ACX since alternate interior angles are congruent. In addition, AC¯¯¯¯¯¯¯¯≅CA¯¯¯¯¯¯¯¯.

Hence, △AXC≅△CBA by The ASA Congruence Postulate. AB¯¯¯¯¯¯¯¯≅XC¯¯¯¯¯¯¯¯ since corresponding parts of congruent triangles are congruent. Since C is the midpoint of XY¯¯¯¯¯¯¯¯, we have XC=CY and XY=XC+CY=XC+XC=2AB by segment addition and substitution.

So, 2AB=XY and AB=12XY. ⧫

Example 1

Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.

M, N and O are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find

A. MN. B. The perimeter of the triangle △XYZ. A. Since O is a midpoint, we have XO=5 and XY=10. By the theorem, we must have MN=5. B. By the Midsegment Theorem, OM=3 implies that ZY=6; similarly, XZ=8, and XY=10. Hence, the perimeter is 6+8+10=24.

We can also examine triangles where one or more of the sides are unknown.

Example 2

Use the Midsegment Theorem to find the value of x in the following triangle having lengths as indicated and midsegment XY¯¯¯¯¯¯¯¯.

By the Midsegment Theorem we have 2x−6=12(18). Solving for x, we have x=152.

Lesson Summary

8 0

3 years ago

A line segment has endpoints at (-4,-6) and (-6,4). Which reflection will produce an image with endpoints at (4,-6) and (6,4)?

GrogVix [38]

The reflection that mapped line segment given by:
(-4,-6) and (-6,4)
to
(4,-6) and (6,4)
is a reflection on the y-axis which follows the rule:
(x,y)→(-x,y)

5 0

3 years ago