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11111nata11111 [884]
3 years ago
12

How to find the length of the hypotenuse correct to one decimal place​

Mathematics
1 answer:
hichkok12 [17]3 years ago
5 0

Answer:

If you are provided with the lengths of two sides a triangle, you can find the third side using the Pythagorean Theorem: a^2+b^2=c^2.

Variable c represents the length of the hypotenuse, and variables a and b represent the lengths of the other two sides.

Please mark as Brainliest! :)

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14. The prime factorization of a number is
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Answer:

See below.

Step-by-step explanation:

The number = 5 * 7 * 23.

5  * 7 = 35 must therefore also be a factor.

If we divide the number by 35 we get 23!.

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(2x-*y-6)-(-7x*y-2)<br> Multiply then simplify
ladessa [460]

Answer:

62652

Step-by-step explanation:

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Pls can any of you help with this.​
Lorico [155]
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What is 3p + 2 &gt; -10
bezimeni [28]

Answer:

p> -4

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2 years ago
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

t=-\text{ln}(\frac{1}{3})\cdot 20

t=-(\text{ln}(1)-\text{ln}(3))\cdot 20

t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

6 0
3 years ago
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