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tia_tia [17]
1 year ago
5

Pls help me solve this question, also include step by step calculation ya. Thks

Mathematics
1 answer:
sp2606 [1]1 year ago
6 0

\displaystyle\\Answer:\ x=\sqrt{2}\ \ \ \ \ x=-\sqrt{2}\ \ \ \ \ x=\frac{ln5}{2}

Step-by-step explanation:

5x^2-x^2e^{2x}+2e^{2x}=10

Let:\ x^2=t  \ \ \ \ \ e^{2x}=v\\Hence,\\5t-tv+2v^2=10\\5t-tv+2v^2-10=10-10\\5t-tv-10+2v^2=0\\t(5-v)-2(5-v)=0\\(5-v)(t-2)=0\\5-v=0\\5-v+v=0+v\\5=v\\5=e^{2x}

<em>Prologarithmize both parts of the equation:</em>

ln5=lne^{2x}\\ln5=2x*lne\\ln5=2x*1\\ln5=2x\\

<u><em> Divide both parts of the equation by 2:</em></u>

\displaystyle\\x=\frac{ln5}{2}

t-2=0\\t-2+2=0+2\\t=2\\x^2=2\\x=\sqrt{2} \\x=-\sqrt{2}

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im so sorry. U_U *tears fall*

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3 years ago
Please answer this question asap will be marked brainliest
Papessa [141]

Answer: 9.85 inches

Step-by-step explanation:

Each point on the graph represents the total rainfall in one month in this desert. For instance, the 2 points on top of 1.0 inches indicates that there were two months in the year where rainfall was 1.0 inches.

The total rainfall in this desert was therefore:

= 0.4 + 0.5 + 0.5+ 0.6 + 0.85 + 0.85 + 0.95 + 1.0 + 1.0 + 1.05 + 1.05 + 1.1

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4 0
3 years ago
A footbridge is 5 feet wide and is built with wood planks that are 6 inches wide. How many planks wide is the path?
Ede4ka [16]

Here we have to know the conversion between feet and inches.

1 Feet = 12 inches (This is given and is to be known)

So the 5 feet wide footbridge when converted to inches would be about 5*12 = 60 inches

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Each wooden plank is 6 inches,

So the number of wooden planks that are required to build the foot bridge of 60 inches = 60/6 = 10 planks

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7 0
3 years ago
Towers A and B are located 5 miles apart. A ranger spots a fire at a 42-degree angle from tower A. Another fire ranger spots the
lord [1]

Answer: The fire is 3.5 miles from tower B

Step-by-step explanation: Please refer to the attached diagram. The triangle in the attached diagram illustrates the clues given in the question. Both rangers are standing at points A and B respectively with a distance of 5 miles between them, which is line AB. Also, one ranger spots a fire from a tower at an angle of 42 degrees, which is point A. Another ranger spots the same fire from another tower, but from an angle of 64 degrees, which is point B. The fire is at point C on the triangle. Now we have a triangle with only one side known (5 miles) and three angles known (the third angle is computed as 180 - {64+42} which equals 74) which are 64 degrees, 42 degrees and 74 degrees.

The distance from the fire to tower B is calculated using the law of sines. (Note that this is not a right angled triangle, hence we cannot use trigonometric ratios). The law of sines is expressed as follows;

a/SinA = b/SinB or

a/SinA = c/SinC

Depending on the sides and angles we are given and the ones we are to calculate.

The distance from the fire to tower B is line BC, labeled as a in our diagram. Using the law of sines

a/SinA = c/SinC

(Note also that a is directly facing angle A, c is directly facing angle C, and so on)

a/SinA = c/SinC

a/Sin 42 = 5/Sin 74

By cross multiplication we now have

a (Sin 74) = 5 (Sin 42)

Divide both sides of the equation by Sin 74 and we now arrive at

a = 5 (Sin 42)/Sin 74

a = 5 (0.6691)/0.9613

a = 3.3455/0.9613

a = 3.4802

{rounded to the nearest tenth of a mile, a equals 3.5}

Therefore the distance from tower B to the fire is approximately 3.5 miles

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3 years ago
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Vsevolod [243]
-x^2-81=-(x^2+81)=-(x^2-(9i)^2)=-(x+9i)(x-9i)=(-x-9i)(x-9i)\to C.
8 0
4 years ago
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