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Same here. It is 30 degrees something Fahrenheit today and it is going to snow tomorrow. However, I don't think we are going to get that much snow... 

would you be able to find the greatest numerical value of a? 

1)-3

2)2

3)-2

4)1

5)-1

Board games

We can prove this

a + c  = d

-d + c  = d

c = 2d

And

c + d  = b

2d + d = b

3d = b

So....using the original equations with the substitutions...

a + b  =  c  ⇒   -d + 3d =  2d

a + c  = d ⇒  -d + 2d  =  d 

c + d  = b  ⇒  2d + d  = 3d

is -d the largest value of a

a + b  = c       ⇒    b  = c - a    (1)

a + c  = d

c + d  = b    (3)

Equating  (1) and (3)

c + d  =  c - a

d  = - a

So

a  =  -d

I can do the first one

2.23 x 10^17   x       1 mole

                               ________________   =  

                                6.0221415 × 10^23

2.23  x 10^17

_______________   moles  =

6.0221415 × 10^23

0.3703001664773237  x  10^(17 - 23)  ≈  .37 x 10^(-6)  =  3.7 x 10^(-7)  moles

Thank you!!

"C"

sin 35°   [ 1st quadrant angle....all trig functions are positive in this quadrant ]

cos 145°  is negative   [  2nd quadrant angle....cosine is negative in this quadrant ]

Don't see an equation but we can model the situation

We have the points  (8, 9)  and  ( 12,11)

The slope between these points is   [ 11- 9 ] / [ 12- 8]  =  2/ 4   =  1/2

So....using the point  (8,9)  and the slope, we have

y =  (1/2)(x - 8) + 9

y  = (1/2)x - 4 + 9

y  = (1/2)x +  5

x is a central angle that intercepts the same arc as the 53°  angle

So....its measure is twice as much   =  106°

y intercepts  an arc of   360 - 106   =  254°

And since y is an inscribed angle intercepting this arc, its measure  is 1/2 of the arc's measure = 127° 

We just take the cube root of this number, @Guest. Solving, we get \(\sqrt[3]{216}\) or \(\boxed{6}\) centimeters. 

Don't know what  angle "AC" is...but if you want to know  the angles, we have

Because BD  = AD.....then angle ABD  = angle BAD.....let these angle measures  = x

And because  AC = AD....then angle ADC = angle ACD.....let these angle measures = y

In triangle ABD, using the Exterior Angle Theorem,  angle ADB  =  x + y

In triangle DAC, angle DAC  =  angle BAD  = x

So....we have the following system.....

x + x + x + y   = 180   ⇒    3x + y  =  180   ⇒   -6x - 2y  = -360   (1)

x + 2y  = 180    (2)

Add (1)  and (2)   and we get that

-5x  = -180

x  = 36°

And using (2)

36 + 2y  = 180

2y  = 144

y = 72°

So   angles  ABD, ADB, DAC   = 36°

And angles ADC, ACD   = 72°

And angle ADB  = 36 + 72   =   108°

The inradius of a polygon can be found by \(A=rs\), where \(A\) is the area, \(r\) is the inradius, and \(s\) is the semi-perimeter. We know that the area is \(\frac{1}{2}*6*8=24\), based on the information given, and \(\frac{6+8+10}{2}=12\) which is considered the semi-perimeter. Now, we have to solve for the radius, and plugging our values in, we reach \(24=12r, 12r=24, \boxed{r=2}\). 

I agree with @CPhill! We have \(\frac{5a+4a^2}{20}\) after joining \(\frac{a}{4}+\frac{a^2}{5}\), together. Now. we have \(\frac{\frac{5a+4a^2}{20}}{25}\) which is \(\frac{5a+4a^2}{20\cdot \:25}\), or our final answer of \(\frac{5a+4a^2}{500}.\)

A tangent  to a circle  meeting a radius always does so at right angles

So  angle OAB   =  90°

What is   "E"   ???

Is this process confusing???....it might not be the same as what you are presented....but....somehow....I find it easier

tysm!

29)   factor top / bottom

( x + 2) ( x - 2)

___________

-3 ( x + 3)

"Holes"  will only be created when we can "cancel" factors

Here....nothing "cancels"  ....so......no "holes"

We will have  a vertical asymptote at x  = -3    and something known as a "slant asymptote"

The latter will occur because we have a greater degree polynomial divided by a lesser degree polynomial

30)   factor top / bottom

(x + 3) ( x - 2)                    (x + 3) ( x - 2)

______________   =       _____________

3 ( x^2 + 4x + 3)               3(x + 3) ( x + 1)

Note that  the  "x + 3"  factor  will "cancel"

So.....to find the "hole".....set    x +  3  =  0    and solve for x......and we have

x + 3  =   0

x  =  -3

So.....we have a hole at  x  = -3

See the graph, here :  https://www.desmos.com/calculator/oshh1sofob

RP.....If you hold down the left button on your mouse/laptop  and drag the cursor along the curve...note....at  x  = -3.....you will see "undefined"   =  (a "hole" here )

Thanks CPhill! ^-^

Simplify the top

Cross-multiply   and keep the same sign between the terms as we have between the fractions

5a + 4a^2

Take the product of the denominators

4 * 5  =  20

So.....we have

[ 5a + 4a^2]              25

__________  ÷       ___

       20                      1

Just like "regular" fractions......Invert the second fraction and multiply

[ 5a + 4a^2 ]                  1

___________      x      _____

     20                            25

5a  +  4a^2

_________

   500

MA = 20   and MX  = 9

And the perpendicular bisector  of  AB  =  CM

So, triangle MAX will be a right triangle with angle AMX  = 90°

Draw  XN   perpendicular to AC

And AD bisects angle BAC...so....

And angle MAX = angle NAX

And angle AMX = angle ANX  =  90°

And AX  is  a common side to triangles AMX  and ANX

So....by   AAS, triangle  AMX  is congruent to triangle ANX

So.....MX = NX  =  9   =   the distance from X to AC   

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