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Firstly I do not think your simplification is correct.  (Unless I have misunderstood what you have wanted to display)

\(e^{-x}+e^{x+e^{-x}}\\ =e^{-x}+e^xe^{e^{-x}}\\ \displaystyle \lim_{x\rightarrow \infty}\;e^{-x}+e^xe^{e^{-x}}\\ =0+e^\infty*1\\ =e^\infty\\ =\infty\)

My presentation is not great. Limits is not my strong suit.  But still I think the logic is correct.

Here is the graph

Oh ok. Thank you for the answer though! Do you have any recommendations?

A quadratic with roots 1 and -1 could be written as

(x-1)(x+1) = 0

Expand the brackets and we get:

x^2 +0x -1 = 0 

So, the solution is 0 and -1

a - b = 1

a = b + 1

a^3 = (b+1)^3

= b^3 +3b^2 +3b + 1

As mentioned in the question above, a^3 - b^3 = 1

We have proven a^3 = b^3 +3b^2 +3b + 1

So, replacing a^3 with b^3 +3b^2 +3b + 1:

b^3 +3b^2 +3b + 1 - b^3 = 1

Collecting like terms

3b^2 + 3b = 0

3b^2 = -3b

b^2 = -b

b = -1, b = 0

If b = -1:

a-(-1) = 1

a + 1 = 1

a = 0

If b = 0:

a - 0 = 1

a = 1

So if b = -1, a = 0

If b = 0, a = 1

Part (a)

It can be either -1 * 0

Which equals 0

Or 0 * 1

Which equals 0

So the only answer is 0

Part (b)

It can be either -1 + 0

Which equals -1

Or 0 + 1

Which equals 1

So the answers are 1, -1

Part (c)

The possible values for a are 0, 1

The possible values for b are -1, 0

Another solution for this problem is using algebra.

Erwus's solution was probably cleaner, but I might as well post mine.

Let n be the side length of the square

n^2 = 9

(kn)^2 = 25

k^2*n^2 = 25

k^2 * 9 = 25

k^2 = 25/9

k = √25/9

= 5/3 or - 5/3

So 5/3 +(-5/3) = 0

800mL acid + 200mL water = 1000mL of total liquid.

Draining 150mL of total liquid

150/1000 = 15%, which means 15% of the water and 15% of the acid will be drained.

0.85* 800 = 680mL acid

0.85* 200 = 170mL water

= 680mL acid and 170mL water

Adding 150mL water

170 + 150 = 320mL water

So there would be 680mL acid and 320mL water.

Draining 250mL of total liquid

680 + 320 = 1000mL total liquid

250/1000 = 25%, which means 25% of the water and 25% of the acid will be drained.

0.75*680 = 510mL acid

0.75*320 = 240mL water

Adding 250mL water

240 + 250 = 490mL water

So your final answer is 490mL water.

Let's say a = 25% of the season

Let's say n is also 25% of the season.

a=35% win rate

3a/3 = (35*3)/3 % win rate

= 35%

(3a+n)/4 = 50%

3a+n = 200%

If a = 35%, 3a = 105%

105% + n = 200%

n = 95%

So, they must win 95% of the remainder of their games to finish the season with the same number of wins as losses.

I haven't explained this very well, but I hope you understand.

Ask one clear, detailed question at a time to get the most useful answers. Multiple quick questions may get overlooked.

Because the scale factor is negative, triangle A'B'C' is obtained by rotating 180 degrees triangle ABC about point G and then shrinking it. Hence, the overlap between triangles ABC and A'B'C' is the smaller triangle A'BC'. (The region consisting of the intersection of triangle ABC with dilated triangle A'B'C' is actually a kite.) We know that the side lengths of triangle A'B'C' are 3/5 the side lengths of triangle ABC. Since the area of a triangle is proportional to the square of its side lengths, the area of triangle A'B'C' is (3/5)2 times the area of triangle ABC. Therefore, the area of the overlap K is (1−(3/5)2) times the area of triangle ABC. Hence, \begin{align*} \frac{K}{ABC} &= 1-(3/5)^2 \ &= 1 - 9/25 \ &= \boxed{\frac{16}{25}}. \end{align*}

The answer is 16/25.

Since ∠ACB=90∘, we have ∠ACE=30∘. Also, since △DEF is equilateral, ∠DEF=60∘. Then ∠AED=60∘−30∘=30∘. Furthermore, △AED is isosceles since AD=AE, so ∠ADE=30∘. Since the sum of the angles in a triangle is 180∘, we have \begin{align*} \angle AED + \angle ADE + \angle EAD &= 180^\circ \ 30^\circ + 30^\circ + \angle EAD &= 180^\circ \ \angle EAD &= 120^\circ. \end{align*}By the Law of Sines in △ADE, we have \begin{align*} \frac{AD}{\sin \angle EAD} &= \frac{AE}{\sin \angle ADE} \ \frac{AE}{5} &= \frac{AE}{\sin 30^\circ} \ 5 &= \sin 30^\circ \ AE &= \frac{5}{\sin 30^\circ} \ AE &= \boxed{\frac{10}{\sqrt{3}}}. \end{align*}

So the answer is AE = 10/sqrt(3).

Squaring the first equation gives [(x+y)^2=x^2+2xy+y^2,]so 102=32+2xy or 2xy=26. Dividing both sides of x+y=10 by 2, we get x/2+y/2=5, so x+y=10. Thus, (x,y) lies on the line y=10−x. Also, since 2xy=26, xy=13, so (x,y) lies on the hyperbola xy=13. The solutions are (x,y)=(2,8), (−2,12), (4,6), (−4,14).

Therefore, the ordered pairs (x,y) are (2,8),(4,6),(8,2),(12,−2),(14,−4)​

The quadratic has nonreal roots if and only if its discriminant is negative.

The discriminant of the quadratic ax2+bx+c is b2−4ac.

The discriminant of the quadratic 12x2+kx+27 is k2−4⋅12⋅27=k2−1296.

Therefore, the quadratic has nonreal roots if and only if k2−1296<0.

This inequality is equivalent to (k−36)(k+36)<0.

Therefore, the quadratic has nonreal roots if and only if k<−36 or k>36.

In interval notation, the solution is:

k \in (-\infty,-36) \cup (36,\infty)

The scale factor of a dilation is equal to the ratio of the length of the image of a segment to the length of the pre-image of that segment.

Since D is the midpoint of BC, the scale factor of the dilation is the ratio of the length of DE to the length of BC. Since the length of DE is half the length of BC, the scale factor of the dilation is 1/2.

Nevermind. Nothing here.

a - b = 1

a = b + 1

a^3 = (b+1)^3

= b^3 +3b^2 +3b + 1

As mentioned in the question above, a^3 - b^3 = 1

We have proven a^3 = b^3 +3b^2 +3b + 1

So, replacing a^3 with b^3 +3b^2 +3b + 1:

b^3 +3b^2 +3b + 1 - b^3 = 1

Collecting like terms

3b^2 + 3b = 0

3b^2 = -3b

b^2 = -b

b = -1, b = 0

If b = -1:

a-(-1) = 1

a + 1 = 1

a = 0

If b = 0:

a - 0 = 1

a = 1

So if b = -1, a = 0

If b = 0, a = 1

Part (a)

It can be either -1 * 0

Which equals 0

Or 0 * 1

Which equals 0

So the only answer is 0

Part (b)

It can be either -1 + 0

Which equals -1

Or 0 + 1

Which equals 1

So the answers are 1, -1

Part (c)

The possible values for a are 0, 1

The possible values for b are -1, 0

x^2 - 8x + 15 + x^2 - 4x + 3

Collect like terms

= 2x^2 - 12x + 18 

= 2x^2 - 6x + 18 - 6x

= 2x(x-3) - 6(x-3)

= (2x-6)(x-3)

2x = 6

x=3

or x-3 = 0

x=3

So the only solution is x = 3