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3 years ago

I need to know what’s the the length

Mathematics

1 answer:

Monica [59]3 years ago

8 0

Answer:

b=4

Step-by-step explanation:

P.T.c^2=a^2 +b^2

5^2=3^2 +b^2

25=9+b^2

b^2 =25-9

b^2 =16

b=sqrt(16)

b=4

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PYTHAGORAS THEROM AND TRIGONOMETRY RATIO

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Answer:

1) ΔACD is a right triangle at C

=> sin 32° = AC/15

⇔ AC = sin 32°.15 ≈ 7.9 (cm)

2) ΔABC is a right triangle at C, using Pythagoras theorem, we have:

AB² = AC² + BC²

⇔ AB² = 7.9² + 9.7² = 156.5

⇒ AB = 12.5 (cm)

3) ΔABC is a right triangle at C

=> sin ∠BAC = BC/AB

⇔ sin ∠BAC = 9.7/12.5 = 0.776

⇒ ∠BAC ≈ 50.9°

4) ΔACD is a right triangle at C

=> cos 32° = CD/15

⇔ CD = cos32°.15

⇒ CD ≈ 12.72 (cm)

Step-by-step explanation:

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3 years ago

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An 8-sided die is rolled. Find the probability of rolling a 1 or even number

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The probability of rolling a 1 or an even number is 5/8.

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3 years ago

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Step-by-step explanation:

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Answer:

Exercise 1: The length of the unknown leg is 4 inches.

Exercise 3: The following straws can be used to construct the triangle: b) $3\,in$ , c) $4.5\,in$ , d) $6.5\,in$ , e) $10\,in$ , f) $13.5\,in$

Exercise 4: Possible options of this exercise: 1) (2, 4, 5), 2) (4, 5, 6), 3) (5, 6, 10), 4) (1, 2, 11), 5) (1, 4, 11), 6) (1, 10, 11)

Step-by-step explanation:

Exercise 1:

Let suppose that triangle represented in the figure is a right triangle, the length of the missing leg is determined by Pythagorean Theorem:

$y = \sqrt{l^{2}-x^{2}}$ (1)

Where:

$l$ - Hypotenuse, in inches.

$x$ - Known leg, in inches.

$y$ - Unknown leg, in inches.

If we know that $l = 11\,in$ and $x = 10\,in$ , then the length of the unknown leg is:

$y = \sqrt{21}$

Since 4 is the least whole number closest to $\sqrt{21}$ , then we conclude that the length of the unknown leg is 4 inches.

Exercise 3:

The range of possible lengths for the missing side of the triangle is represented by the following simultaneous inequality:

$x + y > l > x-y$ (2)

Where:

$x$ - Greater side, in inches.

$y$ - Lesser side, in inches.

$l$ - Missing side, in inches.

If we know that $x = 8\,in$ and $y = 6\,in$ , then we have the following range of missing sides:

$14\,in > l > 2\,in$

The following straws can be used to construct the triangle: b) $3\,in$ , c) $4.5\,in$ , d) $6.5\,in$ , e) $10\,in$ , f) $13.5\,in$

Exercise 4:

Let check each pair to determine possible constructions by means of the inequality used in Exercise 3:

(i) $x = 4\,in, y = 2\,in$

$6\,in>l>2\,in$

Possible choices: 5 inches.

(ii) $x = 5\,in, y = 2\,in$

$7\,in > l > 3\,in$

Possible choices: 4 inches, 6 inches.

(iii) $x = 6\,in, y = 2\,in$

$8\,in > l > 4\,in$

Possible choices: 5 inches, 6 inches.

(iv) $x = 10\,in, y = 2\,in$

$12\,in > l > 8\,in$

Possible choices: None.

(v) $x = 5\,in, y = 4\,in$

$9\,in > l > 1\,in$

Possible choices: 2 inches, 6 inches.

(vi) $x = 6\,in, y = 4\,in$

$10\,in > l > 2\,in$

Possible choices: 5 inches.

(vii) $x = 10\,in, y = 4\,in$

$14\,in > l > 6\,in$

Possible choices: None.

(viii) $x = 6\,in, y = 5\,in$

$11\,in > l > 1\,in$

Possible choices: 2 inches, 4 inches, 10 inches.

(ix) $x = 10\,in, y = 5\,in$

$15\,in > l > 5\,in$

Possible choices: 6 inches.

(x) $x = 10\,in, y = 6\,in$

$16\,in > l > 4\,in$

Possible choices: 5 inches.

Possible options of this exercise: 1) (2, 4, 5), 2) (4, 5, 6), 3) (5, 6, 10), 4) (1, 2, 11), 5) (1, 4, 11), 6) (1, 10, 11)

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