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nataly862011 [7]
3 years ago
12

Show how you solved this by each step.

Mathematics
1 answer:
brilliants [131]3 years ago
6 0

Football is green, so 42% of the people like football.

210 people like football.

To find the total number of people surveyed, divide 210 by 42%:


210 / 0.42 = 500


500 people were surveyed.

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Question 8(Multiple Choice Worth 1 points) (07.01 MC) Find the measure of arc DF. Circle A with chords EF and CD that intersect
tatiyna

Answer:   90°

Step-by-step explanation:

As known ∡EGC=(arcEC+arcDF)/2

arcEC+arcDF=70°*2

5x+10+11x+2=140

16x+12=140

16x=128

x=128:16

x=8

So arcDF=11*x+2=11*8+2=90°

7 0
3 years ago
Angle x measures 42•. Find measure of angle y.
harina [27]

Answer:

C

Step-by-step explanation:

5 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
The graph of y=−3x+2
Sophie [7]

The attached image is the graph.

3 0
3 years ago
Read 2 more answers
If 6 is added to three quarters of a certain number, the result is 42 . Find the original number​
Basile [38]

Answer:

Step-by-step explanation

Hello!

3x/4 + 6=42

3x/4=42-6 = 36

3x = 36*4

3x = 144

x= 144/3 =48

3/4 of  48  = (48*3)  / 4 =36

36+6=42

the original number​ is 48

4 0
2 years ago
Read 2 more answers
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