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liberstina [14]
2 years ago
9

Anyone know how to do this?

Mathematics
2 answers:
lesya [120]2 years ago
6 0

<u>To solve this, we need to use our trignometric functions</u>;

   ⇒ <em>look at the diagram attached</em>

<em />

<u>Let's examine what the problem wants</u>:

  • hypotenuse (<em>longest side)</em> of the triangle

<u>Let's examine what the problem gives us</u>:

  • side adjacent to the 45-degree angle
  • angle that has measure of 45-degrees

<u>Let's solve</u>:

   sin(45)=\frac{5}{x} \\5=x*sin(45)\\x = \frac{5}{sin(45)} \\x=7.07

<u>Answer: 7.1</u> (<em>rounded to nearest tenth)</em>

<em />

Hope that helps!

Sidana [21]2 years ago
3 0

Answer:

5\sqrt2

Step-by-step explanation:

This is a 45-45-90 triangle so we can find all the side lengths easily.

The ratio between the base and the hypotenuse of a 45-45-90 triangle is 1:\sqrt2, which means for this triangle it would be 5:5\sqrt2. This means the length of the hypotenuse is 5\sqrt2\\, which is x, which is the answer.

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Answer: The number of letters that must be engraved for the costs to be the same is 125.

Step-by-step explanation:

Since we have given that

0.4e+10+15=0.20e+5+25

We need to find the number of letters.

0.4e+25=0.2e+30\\\\0.4e-0.2e=30-25\\\\0.2e=25\\\\e=\dfrac{25}{0.2}\\\\e=125

Hence, the number of letters that must be engraved for the costs to be the same is 125.

8 0
3 years ago
Read 2 more answers
PLEASE ANSWR DUE BY 11:59 P.M TODAY 50 POINTS
hoa [83]

Answer:

1) 1/4

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4) 5/6

Step-by-step explanation:

1) P(Red) = 1/4

4 options out of which 1 is red

2)

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Green --> Black, White, Yellow

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White --> Black, White, Yellow

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5 0
3 years ago
que condiciones debe cumplir las medidas de un triangulo para que puedas asegurar que es un triangulo rectangulo
nirvana33 [79]

Answer:

ENGLISH

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SPANISH TRANSLATION

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Let m represent number of Miles...

Then let C represent the total cost for each ride at a specific number of Miles.

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The values stand independently because the customer can choose a different right and a different mile.

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3 years ago
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