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Inga [223]
3 years ago
13

Find the exact length of the third side. 5 3

Mathematics
2 answers:
kykrilka [37]3 years ago
4 0

Answer

This is a 3,4,5 triangle, so the side is 4

Step-by-step explanation:

Evgesh-ka [11]3 years ago
3 0

Hi there! :)

\large\boxed{b = 4}

Use Pythagorean theorem to solve:

a² + b² = c² where "c" is the hypotenuse or the diagonal:

Plug in the given values:

(3)² + b² = (5)²

9 + b² = 25

Subtract 9 from both sides:

b² = 16

Take the square root of both side:

√b² = √16

b = 4.

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labwork [276]
Simple.....

using \frac{ y_{1}- y_{2}  }{ x_{1}- x_{2}  }

--->>>

Take any of the points on the line-->> (0,0) and (15,5)

\frac{0-5}{0-15} = \frac{-5}{-15} = \frac{1}{3}

This is your slope-->>\frac{1}{3}

To check it..go one up then three over...does it hit the coordinates? Yes.

Thus, your answer, D.
3 0
3 years ago
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If tanh x = 12 13 , find the values of the other hyperbolic functions at x. sinh x = 12/5 correct: your answer is correct. cosh
Rom4ik [11]
Tanh x  = 12/13

1) Calculate cosh x:

cosh² x = 1/(1-tanh² x) →→ 1/[1-(12/13)²] = 1/(1-144/169) = 1/[(169-144)169]

cosh²x = 169/25 and cosh x = + or - 13/5

2) Calculate sinh x
 
sinh x/cosh x = tanh x →→sinh x = (cosh x)(tanh x)

sinh x = + or - (13/5)(12/13) →→sinh x= + or - 12/5

From here onward you can calculate easily coth x, sech x, csch x

7 0
3 years ago
Find the sum and express it in simplest form (-y^3-y+1)+(-7y^3-6y-5)
Katen [24]

Answer:

( -  {y}^{3}  - y + 1) + ( - 7 {y}^{3}  - 6y - 5) \\  = ( - 1 - 7) {y}^{3}  - (1 + 6)y + (1 - 5) \\  =  - 8 {y}^{3}  - 7y - 4

4 0
3 years ago
There is 2 questions. PLEASE HELP ME
attashe74 [19]
B and c I believe ..........
4 0
3 years ago
1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than
zaharov [31]

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 3, n = 8

$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)

           $=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)

           $=\left(\frac{81}{11} , \frac{-43}{11}\right)

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 7, n = 1

$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)

           $=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)

           $=\left(\frac{136}{8} , \frac{-12}{8}\right)

C(x, y) = (17, –1.5)

8 0
4 years ago
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