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Yakvenalex [24]
2 years ago
9

A sports team donates 252 tickets to a sporting event, with tickets to be shared equally among 9 classrooms. How many tickets do

es each classroom receive?
Mathematics
1 answer:
Tcecarenko [31]2 years ago
8 0

Answer:

28 tickets

Step-by-step explanation:

252/9=28

You might be interested in
Example of a rational number that is a square root
ipn [44]

Answer:

The list of prime number would probably start with 1 ( 2, 3, 5,7)

4 0
3 years ago
Suppose B is a proper subset of C,
Tomtit [17]

Answer:

Maximum: 7

Minimum: 0

Step-by-step explanation:

A proper subset B of a set C, denoted B\subset C, is a subset that is strictly contained in C and so necessarily excludes at least one member of C.

This means that the number of elements in B must be at least 1 less than the number of elements in C. If the number of elements in C is 8, then the maximum number of elements in B can be 7.

The empty set is a proper subset of any nonempty set. Hence, the minimum number of elements in B can be 0.

3 0
3 years ago
Decide if the following statement is valid or invalid. If two sides of a triangle are congruent then the triangle is isosceles.
Naya [18.7K]

Answer:

Step-by-step explanation:

Properties of an Isosceles Triangle

(Most of this can be found in Chapter 1 of B&B.)

Definition: A triangle is isosceles if two if its sides are equal.

We want to prove the following properties of isosceles triangles.

Theorem: Let ABC be an isosceles triangle with AB = AC.  Let M denote the midpoint of BC (i.e., M is the point on BC for which MB = MC).  Then

a)      Triangle ABM is congruent to triangle ACM.

b)      Angle ABC = Angle ACB (base angles are equal)

c)      Angle AMB = Angle AMC = right angle.

d)      Angle BAM = angle CAM

Corollary: Consequently, from these facts and the definitions:

Ray AM is the angle bisector of angle BAC.

Line AM is the altitude of triangle ABC through A.

Line AM is the perpendicular bisector of B

Segment AM is the median of triangle ABC through A.

Proof #1 of Theorem (after B&B)

Let the angle bisector of BAC intersect segment BC at point D.  

Since ray AD is the angle bisector, angle BAD = angle CAD.  

The segment AD = AD = itself.

Also, AB = AC since the triangle is isosceles.

Thus, triangle BAD is congruent to CAD by SAS (side-angle-side).

This means that triangle BAD = triangle CAD, and corresponding sides and angles are equal, namely:

DB = DC,

angle ABD = angle ACD,

angle ADB = angle ADC.

(Proof of a).  Since DB = DC, this means D = M by definition of the midpoint.  Thus triangle ABM = triangle ACM.

(Proof of b) Since angle ABD = angle ABC (same angle) and also angle ACD = angle ACB, this implies angle ABC = angle ACB.

(Proof of c) From congruence of triangles, angle AMB = angle AMC.  But by addition of angles, angle AMB + angle AMC = straight angle = 180 degrees.  Thus 2 angle AMB = straight angle and angle AMB = right angle.

(Proof of d) Since D = M, the congruence angle BAM = angle CAM follows from the definition of D.  (These are also corresponding angles in congruent triangles ABM and ACM.)

QED*

*Note:  There is one point of this proof that needs a more careful “protractor axiom”.  When we constructed the angle bisector of BAC, we assumed that this ray intersects segment BC.  This can’t be quite deduced from the B&B form of the axioms.  One of the axioms needs a little strengthening.

The other statements are immediate consequence of these relations and the definitions of angle bisector, altitude, perpendicular bisector, and median.  (Look them up!)

Definition:  We will call the special line AM the line of symmetry of the isosceles triangle.  Thus we can construct AM as the line through A and the midpoint, or the angle bisector, or altitude or perpendicular bisector of BC. Shortly we will give a general definition of line of symmetry that applies to many kinds of figure.

Proof #2 (This is a slick use of SAS, not presented Monday.  We may discuss in class Wednesday.)

The hypothesis of the theorem is that AB = AC.  Also, AC = AB (!) and angle BAC = angle CAB (same angle).  Thus triangle BAC is congruent to triangle BAC by SAS.

The corresponding angles and sides are equal, so the base angle ABC = angle ACB.

Let M be the midpoint of BC.  By definition of midpoint, MB = MC. Also the equality of base angles gives angle ABM = angle ABC = angle ACB = angle ACM.  Since we already are given BA = CA, this means that triangle ABM = triangle ACM by SAS.

From these congruent triangles then we conclude as before:

Angle BAM = angle CAM (so ray AM is the bisector of angle BAC)

Angle AMB = angle AMC = right angle (so line MA is the perpendicular bisector of  BC and also the altitude of ABC through A)

QED

Faulty Proof #3.  Can you find the hole in this proof?)

In triangle ABC, AB = AC.  Let M be the midpoint and MA be the perpendicular bisector of BC.

Then angle BMA = angle CMA = right angle, since MA is perpendicular bisector.  

MB = MC by definition of midpoint. (M is midpoint since MA is perpendicular bisector.)

AM = AM (self).

So triangle AMB = triangle AMC by SAS.

Then the other equal angles ABC = ACB and angle BAM = angle CAM follow from corresponding parts of congruent triangles.  And the rest is as before.

QED??

8 0
2 years ago
A.) write an equation in slope intercept form for the total cost of any number of tickets at 7 tickets for $5.
marta [7]
Slope intercept form is y = mx + b

7 tickets for $5 means that 1 ticket costs 5/7 dollars

So for part a, the slope intercept form would be:

y = \frac{5}{7}x

where x is the number of tickets, and y is the price.

For part b)

The slope intercept form is:
y = 25x

where x is the number of all you can ride wristbands you buy, and y is the total price.
3 0
3 years ago
Three landmarks of baseballachievement are Ty Cobb’s batting average of 0.420 in 1911,Ted Williams’s 0.406 in 1941, and George B
harina [27]

Answer:

Ted Williams has the highest standardized score, hence Williams is the best of the three

Cobb is the second highest and hence, the better player

George Brett is third

Step-by-step explanation:

Given the data:

Decade : 1910 __ 1940 ____ 1970

Mean, μ: 0.266 __0.267 ___ 0.261

S/dev, σ : 0.0371 _ 0.0326 __ 0.0317

To compute the standard unit for batting averages, we obtain the standardized score for each of Cobb,Williams, and Brett.

Standardized score (Zscore) : (x - μ) / σ

Cobb, x = 0.420

Ted Williams, x = 0.406

George Brett, x = 0.390

COBB:

Zscore = (0.420 - 0.266) / 0.0371 = 4.1509

Ted Williams :

Zscore = (0.406 - 0.267) / 0.0326 = 4.2638

George, Brett = (0.390 - 0.261) / 0.0317 = 4.0694

Ted Williams has the highest standardized score, hence Williams is the best of the three

Cobb is the second highest and hence, the better player

George Brett is third

6 0
2 years ago
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