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Tatiana [17]
3 years ago
11

Mohammad would like to graph the line that represents the total number of baseball cards he will have collected, T, after m mont

hs, given that he buys 5 cards each month and he started with no cards. He would like the graph to show the number of cards after 6 and 12 months. Select from the drop-down menus to correctly complete each statement. To best show this information, the scale for the T-axis of his graph should go from 0 to at least , and the m-axis of his graph should go from 0 to at least .

Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0

Solution:

Number of Baseball cards bought by Mohammad each month = 5

As number of baseball cards is represented by T , which are

T = 0,5,10,15,20,25,30,35,40,45,50,55,60   →→  Y axis

Number of months(m) from which Mohammad started buying cards is given as

m = 0,1,2,3,4,5,6,7,8,9,10,11,12   →→     X axis

The given information represented in equation form

T = 5 m

This is a equation of line in two variable passing through origin.

Slope of line = 5 =Amount of baseball card bought each month

Number of cards bought after 6 months = 5 × 6=30

Number of cards bought after 12 months = 5 ×12 =60

Graph is depicted below.

The scale for the T-axis of his graph should go from 0 to at least 60 , and the m-axis of his graph should go from 0 to at least 12 .


mylen [45]3 years ago
3 0
T-axis of his graph should go from 0 to at least 60 <span>and the </span>m<span>-axis of his graph should go from 0 to at least 12.

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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

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Line AM is the altitude of triangle ABC through A.

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

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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).

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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?)

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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??

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