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Gnesinka [82]
3 years ago
11

Write a story or situation that can be represented by the given equation. 4a+5=12a−11

Mathematics
2 answers:
Len [333]3 years ago
5 0

Answer:

a=4

Step-by-step explanation:

4a+12a=  11

Leto [7]3 years ago
3 0

Answer:

          a = 4

Step-by-step explanation:

STEP 1)  4a - 16  =   4 • (a - 4)

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s,h,q,j,o,l,m...
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Marlon and Michelle have to graph a function with the equation y=x2and are having to debate about which is the most important el
Usimov [2.4K]

Answer:

Correct answer is the domain.

Step-by-step explanation:

A domain is the input set of a function.

A range is the output set of the function.

It is more  important to know the domain of the function to be able to determine the corresponding output set so the function can be graphed. To start with the domain is very useful because every domain element has a corresponding unique output element.

Suppose you started with the range element of 9. The input set for an output of 9 is {-3,3}. This makes it hard to match up the elements. This example highlights why it is important to start with the domain rather than the range.

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 function is defined by ()= −2+6. What is the value of (10)?
kirill115 [55]

Answer:

x

Step-by-step explanation:

4 0
3 years ago
Sketch graphs of these equations:<br><br><br><br> y=x
Vilka [71]

Answer:

wats the equation?

Step-by-step explanation:

4 0
3 years ago
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