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Firlakuza [10]
10 months ago
10

What is the slope of the line represented by 4x-2y=10?

Mathematics
1 answer:
Musya8 [376]10 months ago
5 0

ANSWER:

2

STEP-BY-STEP EXPLANATION:

We have the following equation of line:

4x-2y=10

The equation of the line in its slope and intercept form is like this:

\begin{gathered} y=mx+b \\ \text{where, m is the slope and b is the y-intercept} \end{gathered}

Therefore, we must solve for y to know the value of the slope (m), like this:

\begin{gathered} 4x-2y=10 \\ 2y=4x-10 \\ y=\frac{4x-10}{2} \\ y=\frac{4x}{2}-\frac{10}{2} \\ y=2x-5 \\ \text{therefore,} \\ m=2 \end{gathered}

The slope is 2

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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??

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In how many ways can you choose 2 side dishes out of 8
Alina [70]
The first choice can be any one of the 8 side dishes.
               For each of these . . .
The 2nd choice can be any one of the remaining 7.

     Total number of ways to pick 2 out of 8 = (8 x 7) = 56 ways .

BUT ...

That doesn't mean you can get 56 different sets of 2 side dishes.

For each different pair, there are 2 ways to choose them . . .
(first A then B), and (first B then A).  Either way, you wind up with (A and B).

So yes, there are 56 different 'WAYS' to choose 2 out of 8.
But there are only 28 different possible results, and 2 'ways'
to get each result.
4 0
2 years ago
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I really need help on this geometry question, please, and thank you.
photoshop1234 [79]

Answer:

250

Step-by-step explanation:

you have to use the fact that the big triangle and the small triangle are similar and write the proportions

\frac{125}{10} =\frac{AB}{10}  we took half of the side because of ΔABD and ΔBD..

AB= (125*10)/10=125

is AB =125 then AC is duble AC= 2*125=250

7 0
2 years ago
Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?
Bogdan [553]

Step-by-step explanation:

\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}

\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}

\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}

\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}

\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}

\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}

Taking sin²θ common in both numerator & denominator, We get :

\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}

\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}

\mathsf{\implies \dfrac{1 -cot^2\theta}{1 + cot^2\theta}}

\mathsf{Given :\;cot\theta = \dfrac{1}{\sqrt{5}}}

\mathsf{\implies \dfrac{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2}{1 + \left(\dfrac{1}{\sqrt{5}}\right)^2}}

\mathsf{\implies \dfrac{1 - \dfrac{1}{5}}{1 + \dfrac{1}{5}}}

\mathsf{\implies \dfrac{\dfrac{5 - 1}{5}}{\dfrac{5 + 1}{5}}}

\mathsf{\implies \dfrac{5 - 1}{5 + 1}}

\mathsf{\implies \dfrac{4}{6}}

\mathsf{\implies \dfrac{2}{3}}

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