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Elan Coil [88]
3 years ago
7

How to solve this Math equation

Mathematics
2 answers:
m_a_m_a [10]3 years ago
7 0
The answer will be 7.761!
4vir4ik [10]3 years ago
4 0

This is how. (Below)

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Exponents: Can me help me? This is the one, I just can't get.
aliina [53]

Answer:

1

Step-by-step explanation:

x^y * x^-y

When the bases are the same,   we add the exponents when multiplying

x^(y-y)

x^0

Anything to the zero power is 1  (except 0)

1


4 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
Choose the rate of change (slope) to match the line.
V125BC [204]
Bbbbbbbbbbbbbbbbbbbbbb
8 0
3 years ago
Please help!! This topic is incredibly confusing and assistance is very much needed & appreciated! Thanks :)
Artyom0805 [142]
I got 8/10 or 80% because if you add up all sandwichs together, you get 50, then you add up ham and tuna sandwichs and get 40 so 40/50 would be 80/100 to make it a percent or fraction.. hopefully this makes sense and i helped:) i wish i could just send the work to you so it could make more sense lol
7 0
3 years ago
Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of R by

\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
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