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Drupady [299]
3 years ago
11

OMG PLEASE Help me !!!! ILL GIVE YOU BRAINLIEST !!! Explain in order to get it

Mathematics
1 answer:
Doss [256]3 years ago
4 0

Answer:

I'm not sure but I believe the correct answer is line L2.

Step-by-step explanation:

It seems to be closer to more points than the other lines.

If it is not L2, it must be L3.

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A triangle formed by three squares joined at their vertices. The side lengths of the squares are 2, 3 1/3, and 2 2/3. Is the tri
S_A_V [24]
If it is a right triangle, then this must be true:

(2)^2 + (2\frac{2}{3})^2 = (3\frac{1}{3})^2

(2)^2 + (\frac{8}{3})^2 = (\frac{10}{3})^2\\\\4 + \frac{64}{9} = \frac{100}{9}\\\\\frac{36}{9} +\frac{64}{9} =\frac{100}{9}\\\\\frac{100}{9}=\frac{100}{9}

The correct answer is: <span>A. Yes, because the areas of the two smaller squares add up to the area of the largest square.</span>
7 0
3 years ago
elena finds that the area of a house on a scale drawing is 25 square inches. the actual area of the house is 2,025 square feet.
zavuch27 [327]
Answer: 2,025/25=81

81 is the scale factor
8 0
4 years ago
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Find the length of the hypotenuse. The measure of the hypotenuse?
Eduardwww [97]

Answer:

Since it's a right angled triangle

Hypotenuse = h

h² = 7²+24²

h= √49 + 576

h = √625

h= 25

Therefore hypotenuse is 25

Hope this helps.

8 0
4 years ago
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Sara rented a bike from Mark's bikes. It cost 14$ plus 3$ per hour. If Sara paid 32$ then she rented the bike for how many hours
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3 years ago
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Solve the differential equation. y' + 5xey = 0.
galben [10]

Answer:

The solution is     y = - ln(\frac{5}{2}x^{2} + C)

Step-by-step explanation:

To solve the differential equation, we will find y

From the given equation, y' + 5xey = 0.

That is, y' + 5xe^{y} = 0

This can be written as

\frac{dy}{dx} + 5xe^{y} = 0

Then,

\frac{dy}{dx} = - 5xe^{y}

\frac{dy}{e^{y}}   = - 5x dx

Then, we integrate both sides

\int {\frac{dy}{e^{y}}}  =\int {- 5x dx}

\int {e^{-y}dy }}  =\int {- 5x dx}

Then,

-e^{-y} = -\frac{5}{2}x^{2} + C

e^{-y} = \frac{5}{2}x^{2} + C

Then,

ln(e^{-y}) = ln(\frac{5}{2}x^{2} + C)

Then,

-y = ln(\frac{5}{2}x^{2} + C)

Hence,

y = - ln(\frac{5}{2}x^{2} + C)

5 0
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