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I think you are missing the graph that shows the graph of the exponential function. 

However, I can make a geuss to what the exponential function is using the two points :). 

Exponential functions are usually in the form of \(y=ax^b\) where a is the y intercept and b is the growth factor. 

With the two points you provided, f(0) = 1 is defintely the y intercept at point (0, 1), meaning the a value is 1. 

My best geuss at what the growth factor is would probably be that the y value doubles everytime x increases by 1, because that would make the most sense for a exponential function.

This would mean the b value or growth factor is 2. 

So, my best geuss without a graph of what f(x) is would probably be \(f(x) = 2^x\). 

I'm not sure if it's right, but it does satisfy the two points given, and that was all, so I hope it was a bit helpful!

Thanks!

So first off, let's write the function in standard form . Warning, this problem is quite straightforward, and you could probably geuss it...maybe

Distributing the 2 into the funcction gets us \(f(x) = -6x+4\)

DOMAIN: 

For the domain of this function, we just need to check whether or not there are any limits to what x can be. 

For example, in the function \(f(x) = \sqrt{2x+54}\), we must have the descriminant as greater or equal to 0, meaning there are restraints on the domain. 

In this function however, there are none, since x could be anything and it would not violate any concepts. 

Domain: All Real Numbers

RANGE:

Now, the range is limited usually because the domain is limited in someway. 

In this case, there are no limited to the domain, so there are no limits to the range. 

Range: All Real Numbers

Yep, both the range and domain are ALL REAL NUMBERS. 

In scientific notation:

Domain: \((-\infty, \infty)\)

Range: \((-\infty, \infty)\)

Thanks!

Very nice, Max  ....this one is  somewhat difficult  !!!

Could you explain how the area is  8x / 2  ???......my small mind doesn't understand (LOL!!!)

Simplifying, we have \((3y - 11)^4\).

So the coefficient of \(y^4\) is \(3^4 = 81\)

Extract the useful information: M is the midpoint of BC. 

Then triangle AMC has half the base of triangle ABC, and the same height. 

The area of triangle AMC would be half that of triangle ABC, so the answer is 90.

Label the diagram as shown. Let \(\angle FCE = \theta\) and \(EC = x\).

Let H be a point on AC such that FH is perpendicular to AC. Then \(FH = 8\).

Note that \(CE = FD = HA = x\). Then \(CH = 10 - x\). 

\(CF = \sqrt{8^2 + (10 - x)^2} = \sqrt{x^2 - 20x + 164}\) by using Pythagoras theorem in \(\triangle CFH\).

In \(\triangle ECG\), \(\sin \theta = \dfrac{EG}{EC} = \dfrac1x\).

In \(\triangle FHC\), \(\sin \theta = \dfrac{FH}{CF} = \dfrac8{\sqrt{x^2 -20x+164}}\).

So we have \(\dfrac1x = \dfrac8{\sqrt{x^2 - 20x + 164}}\). Squaring gives \(\dfrac1{x^2} = \dfrac{64}{x^2 - 20x + 164}\).

Rearranging, 

\(64x^2 = x^2 - 20x + 164\\ 63x^2 + 20x - 164 = 0\\ x = \dfrac{-20 \pm \sqrt{20^2 - 4(63)(-164)}}{2(63)}\\ x = \dfrac{8 \sqrt{163} - 10}{63}\text{ (reject negative root)} \)

Hence, the area is \(\dfrac{8x}2 = \dfrac{32 \sqrt{163} - 40}{63}\).

This is wrong, but I can't edit it, see CPhill's answer below

... and let $X$ be a point such that $MH = MX$, as shown below

Did you forget to attach a diagram?

The nth term  = 1 + 5(n  -1)

The sum of the series =  [ first term + last term ] *  number of terms /  2

We can solve this

[  1  +  1 + 5(n - 1) ] (n / 2)  >   40

[ -3 + 5n ] (n)  >  80 

5n^2 - 3n - 80  >  0        set   =   0

5n^2  -3n - 80  = 0

We need n to be  an  integer

Use the Quad Formula to find n and  take its ceiling

n  =  ceiling  (  [ 3 + sqrt [ 3^2  - 4 * 5 * -80 ] ] / 10 )   =   5

The number that is  called is    1 + 5(n -1)  = 1 + 5 (5 - 1)  =  1 + 5 (4)   =  21          

Suppose O is the centre of the circle. Then \(\Delta OP_1 P_2\) is an isosceles triangle with \(OP_1 = OP_2 = 1\) and \(\angle P_1OP_2 = \dfrac{360^\circ}{10} = 36^\circ\).

Then \(P_1P_2 ^2 = 1^2 + 1^2 - 2(1)(1) \cos 36^\circ = 2 - \dfrac{1 + \sqrt 5}2 = \dfrac{3 - \sqrt 5}2\\ \). Note that \( \left(\dfrac{\sqrt 5 - 1}2\right)^2 = \dfrac{3 - \sqrt 5}2\). Then \(P_1P_2 = \dfrac{\sqrt 5 - 1}2\).

Repeating the same process 10 times gives \(P_1 P_2 + P_2 P_3 + P_3 P_4 + \cdots + P_9 P_{10} + P_{10} P_1 = 10\cdot \dfrac{\sqrt 5 -1}2 = 5(\sqrt 5 - 1)\).

Divide by 5 on both sides.

\(125x = \dfrac 95\)

a2 =  a1 + d

a3 = a1 + 2d

So

a1 + (a1 + d)  =  4      →     2a1 + d =  4            (1)

(a1 + d)  + (a1 + 2d)  =  5   →   2a1 + 3d =  5    (2)

Multiply  (1)  by   -3 and  add to (2)

-4a1  =  -7

a1  = -7 / -4  =  7 / 4

S1  = 5

S2 =  15

Then   

a1 =  5     and   a2  = 10         d =  5

So

a15 =  a1   + (15 - 1) (d)  =    5 + (15 - 1) (5)  =  5 + 14  * 5  =  75

S15  =[ a1 + a15 ] ( 15 / 2)  =    [  5 + 75 ] (15 / 2)  =     600

Honestly, it's not too hard to just list the numbers and sums out, since the sum we're trying to get is quite small. 

So, when Laverne yells out 31, Shirley would scream out yelling!

:) :) :)

Thanks!

The  slant height of the cone =  21

The circumference of the  cone  =  (120 / 360)  * 2 pi * 21  =  14 pi

The  radius of the cone can  be  found as

14 pi  = 2 pi * r

r = 7  

The height of the cone =  sqrt [ l^2 - r^2 ]  = sqrt [21^2 - 7^2 ]

Volume of the  cone  is

(1/3) pi *  r^2 * h   =

(1/3) pi  * (7)^2 * sqrt [ 21^2 - 7^2 ]  ≈  1015.94

I don't think this is possible  if AB  =  10   ( it must be greater)

If we can....let us imagine this  figure ....  { AB  is  actually ≈ 13.45 }

Let DC = DE =  4

Let AB  = "10"

Let AE =  "x"

And BE  =  BC  =   "10 - x"

And AD  =   sqrt [ DE^2 + AE^2]  = sqrt [ 4^2 + x^2 ] 

Note that ...Triangle ADE ≈  Triangle ABC

So

DE / AD  =  BC / AB

4 / sqrt [ 4^2 + x^2  ] =  (10 - x )  /  10        

The only real solution to  this  is  x =   0  =   AE    which is  impossible

To solve the equation EIGHT / FOUR = TWO, where each letter stands for a different digit, we can start by considering the possible values of the letters.

Since we are dividing a four-digit number by a four-digit number and obtaining a three-digit number, we know that E and F must be 9 and 1, respectively, as they are the leading digits in the division.

With this information, let's construct the division:

```
   9 1
-------
   T W O
```

Now, let's consider the possible values for T and W.

Since 91 divided by 4 is not a whole number, T must be greater than 2. Let's try T = 3:

```
   9 1
-------
  3 W O
```

For the division to yield a three-digit result, the number formed by the last two digits of the dividend (1W) must be divisible by 4. Therefore, W must be an even number. The only even digit left is 2.

```
   9 1
-------
  3 2 O
```

Now, let's find the value of O. Since 91 divided by 4 is 22 with a remainder of 3, O must be 3.

```
   9 1
-------
  3 2 3
```

Therefore, the solution to the equation EIGHT / FOUR = TWO, where each letter stands for a different digit, is:

```
   9 1
-------
  3 2 3
```

So, E = 9, I = 1, G = 3, H = 2, T = 9, W = 2, and O = 3.

Angle  ABC  =105

Let AD = x    CD = 10 - x

BC / sin 30  = AC /sin 105

BC /sin30 = 10 /sin105

BC = 5/sin105

BA / sin 45 =  AC / sin 105

BA / sin45 = 10 /sin105

BA = (10/sqrt 2) / sin 105  =   20 / (1 + sqrt 3)

Since BD is a  bisector, then

BA / AD = BC / CD

(10/sqrt2)/ sin 105 / x  = 5/sin105 / (10-x)

(10/sqrt 2) (10-x) / sin 105  =  5x / sin105

(10/sqrt 2) (10 -x)  = 5x

100/sqrt 2  - (10/sqrt 2)x = 5x

100/sqrt 2  = (5 + 10/sqrt 2) x

x = (100/sqrt 2) / ( 5 + 10/sqrt 2)  =   (100) / (5sqrt 2 + 10)  =  20 / (2 + sqrt 2) = AD

[ABD ] = (1/2)BA * AD  * sin 30  =  (1/4) [20 / (1 + sqrt 3)] * [ 20 /(2 + sqrt 2) ]  ≈ 10.72

Let's start with what we should do for any geometry question, which is to of course, graph it. 

Note that we have right triangles ADC, ABE, ABD, BHD, BEC and triangles ABE and ADC are a 30-60-90 and a 45-45-90 respectively. 

Now, it's pretty simple from here. 

We first draw the altitude for triangle ABC from point C to line segment AB, it intersects point H along the way.

Also, since we know BAC is 60 degrees, HCA would be 30 degrees!

30 degrees!

Thanks!

I think this is more of a trick question 

First off, I graphed the triangles. Note, please excuse my awful drawing skills. 

Alright, now we have a graph, we can easily visualulize the angles needed. 

We honestly don't need to do much math if we notice one major fact. 

If we rearrange the order of the angles we need to find a bit, \(\angle ZNM + \angle ZMN + \angle NZM + \angle PNM + \angle PMN+ \angle NPM\) , we notice one big factor. 

The angles \(\angle ZNM + \angle ZMN+ \angle NZM\) are the three angles of the triangle ZMN, meaning they simply add up to 180 degrees. 

The same goes for the angles \(\angle PNM + \angle PMN + \angle NPM\). They are the 3 angles to triangle PMN, meaning they also add up to 180 degrees. 

This means that \(\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM = 180 + 180 = 360\). 

360 degrees is the final answer!

Thanks! :)

That is not correct 

Here is the answer to the second problem.

To find the area of the square inscribed in the right triangle, we can use the fact that the square is inscribed, meaning that its diagonal is also the hypotenuse of the right triangle.

Let the side length of the square be \( s \). Since the square is inscribed in the right triangle, its diagonal is equal to the hypotenuse of the right triangle.

By the Pythagorean theorem, the length of the hypotenuse of the right triangle is given by:

\[ \sqrt{1^2 + 3^2} = \sqrt{10} \]

Since the diagonal of the square is also the hypotenuse of the right triangle, its length is \( \sqrt{10} \).

Now, the diagonal of the square divides it into two congruent right triangles. Each right triangle has legs of length \( s \) and \( s \), and a hypotenuse of length \( \sqrt{10} \).

Using the Pythagorean theorem again, we have:

\[ s^2 + s^2 = (\sqrt{10})^2 \]

\[ 2s^2 = 10 \]

\[ s^2 = \frac{10}{2} \]

\[ s^2 = 5 \]

\[ s = \sqrt{5} \]

So, the side length of the square is \( \sqrt{5} \).

The area of the square is the square of its side length:

\[ \text{Area} = (\sqrt{5})^2 \]

\[ \text{Area} = 5 \]

Therefore, the area of the square inscribed in the right triangle is \( 5 \) square units.

Note that \((\sin (a))^5 < 1\) and \((\cos(a))^5 < 1\).

Then, we have

\(\begin{array}{rcl} (\sin(a))^7 + (\cos(a))^7 &=& (\sin(a))^2(\sin(a))^5 + (\cos(a))^2 (\cos(a))^5\\ &<& (\sin(a))^2 + (\cos(a))^2\\ &=& 1 \end{array}\)

The inequality is shown.

Suppose the radius of the circle = r units 

\(\pi r^2 = 25\pi\\ r = 5\)

Then, the radius of the sphere is \(\sqrt{r^2 + 5^2} = 5 \sqrt 2\) units.

Then the surface area is \(4 \pi (5 \sqrt 2)^2 = 200\pi\).

Using Pythagoras theorem in \(\triangle ABX\) gives \(AX^2 + BX^2 = AB^2\), substituting AX = 15 and AB = 25 gives

\(15^2 + BX^2 = 25^2\\ BX^2 = 400\\ BX = 20\)