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The answer to this question is x=14. 

The lateral  area will be comprised of 8 congruent triangles

The slant height  of each triangle  = sqrt (6^2 + 6^2)  = sqrt (72)  = 6sqrt (2) cm

And the base of each triangle  = 5 cm

So.....the total lateral  area  =

8 * (1/2) (base) (slant height)  =

8 (1/2) (5) (6sqrt(2) ) ≈  169.7 cm ^2

The answer is x=13

The sum of the areas of all 8 triangles is the lateral area.

And, each of these triangles are the same size.

So...

lateral area = 8 * area of one of these triangles

lateral area = 8 * (1/2) * base * height

From the Pythagorean theorem:

6² + 6² = height²          \(\rightarrow \quad \text{height} = \sqrt{72}=6\sqrt2\)

lateral area = 8 * (1/2) * 5 * 6√2

lateral area = 120√2     ≈     169.7   cm²

f(x)  = x^3  +  x

If this is the position function, the velocity function is the derivative of this

f'(x)  =  3x^2  + 1

Assuming that you meant x  = 5, we have that the velocity at that time  = 

f'(5)  =  3(5)^2  +  1   =    76  [ units per second  ??  ]

Just looking at the left half of the front face...

this is a 30º 60º 90º triangle with the side across from the 30º angle = 3 m

So....the side across from the 60º angle = 3\(\sqrt3\)  m

12 = \(\frac67\)x

Multiply both sides of the equation by \( \frac76 \)

12 * \( \frac76 \) = \(\frac67 \)x * \( \frac76 \)

14 = x

y''-4y'+3y=0             y(0)=0 and y'(0)=-1

We can solve this by letting the general solution be

y(t)  =  C₁ e ^r1t  +  C₂ e ^r2t  

And  we can find r1  and r2 thusly

(r  - 3)  ( r - 1)  =  0         setting  both factors to 0 and solving for r  we have that r1   = 3  and r2  = 1

So......the general solution is :

y(t)  =  C₁ e^3t  +  C₂ e^1t      and

y' (t)  = 3C_i e ^3t  +  C₂ e ^1t

Applying the initial conditions, we have that

0  =  C₁  + C₂     →    C₂  =  - C₁     (1)

-1  = 3C₁  + C₂       (2)

Subbing  (1)  into (2)  we have that

-1  =  3C₁  -  C₁

-1  =  2C₁

-1/2  = C₁    →    C₂  = -C₁  =  1/2

So......the actual solution is

y(t)  =  (-1/2) e ^3t   +  (1/2) e^1t

12=6/7x

19 would be your answer

To solve this for x.....

3x + (-4) = 2x + 9

3x - 4 = 2x + 9

Add 4 to both sides of the equation.

3x = 2x + 9 + 4

3x = 2x + 13

Subtract 2x from both sides of the equation.

3x - 2x = 13

x = 13

Thanks 

volume of a cube = (edge length)³     

it would equal 105

I might be wrong here...but I think we need to know something about the time from the information given in the question.

Does "x" stand for time in the formula? Or are you given another equation that tells us the x or y coordinate at any point in time?

CPhill's answer is the CORRECT one:

Distance =1.5S(Mr. Tan's leg) + (1.5*76 +26)(Mr. Sammy's leg) =4S and S =56 Km/h x 4 =224 Km.

3 * 5 * 7  =  105

Thanks 

\(\frac{96}{36}*39=\frac{96*39}{36}=\frac{3744}{36}=104\)

-3x² + 4x + 1 = 16x + 13

Add  3x²  to both sides of the equation.

4x + 1 = 3x² + 16x + 13

Subtract  4x  from both sides of the equation.

1 = 3x² + 12x + 13

Subtract 1 from both sides of the equation.

0 = 3x² + 12x + 12

Now we can use the quadratic formula, with a = 3, b = 12, and c = 12, to solve for x.

\(x = {-12 \pm \sqrt{12^2-4(3)(12)} \over 2(3)} \\~\\ x = {-12 \pm \sqrt{144-144} \over 6} \\~\\ x = {-12 \pm \sqrt{0} \over 6} \\~\\ x = {-12\over 6} \\~\\ x = -2\)

 It operates on an ordered listing, such as integers or alphabet letters. 

 So succ(4) is 5 

 succ(B) is C 

Mr Tan was travelling from Town L to Town M at a uniform speed.He drove past Mr Samy who was travelling at a uniform speed of 76 km/h in the opposite direction.1.5 hours later,Mr Tan reached Town M while Mr Samy was still 26 km away from Town L.Mr Tan took 4 hours to travel from Town L to Town M.Find the distance between the two towns.

Call Mr. Tan's rate, R

The total distance between the  towns  = Mr. Tan's Rate * 4 hrs = 4R....and this equals to

The distance that  Mr. Tan traveled in (1.5) hrs  =   1.5R    plus

The distance that Mr Sammy traveled  in 1.5 hrs  + 26 km  =   1.5(76)  + 26 km

So we have this

4R  =  1.5R  +  1.5 (76)  +  26  simplify

2.5R  =  140    divide both sides by  2.5

R  = 56 km / hr  =   Mr. Tan's Rate

So.....the total distance between the towns  =  4 * (56)  =  224 km

Average speed = Distance / Time
NOTE: After the two men meet, Mr. Tan drives for another 1.5 hours.
Let the distance between the 2 towns = D
Let the average speed of Mr. Tan = S. But we have:
4S = D. And we also have Mr. Sammy driving:
1.5*76 + 26 =140 km - The distance Mr. Tan had to cover in 1.5 hours. Therefore, Mr. Tan's average speed:
140/1.5 =93 1/3 Km/h. And the distance between the 2 towns:
D =4 hours x 93 1/3 Km/h = 393 1/3 Km.

You can just type in " 1/6 " into the bar and it will turn that fraction into a decimal.

1/6 ≈ 0.166667