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olya-2409 [2.1K]
3 years ago
7

AB is tangent to circle O at B. Find the length of the radius r for AB = 9 and AO = 10.5. Round to the nearest tenth if necessar

y. The diagram is not to scale.

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
4 0

Answer:

5.4

Step-by-step explanation:

At the point of tangency, the degree measure is 90. Hence Angle ABO is 90 degrees. So triangle ABO is right triangle.

In a right triangle, we can use the pythagorean theorem, which says:

Leg1^2 + Leg2^2 = Hypotenuse^2

Where

Leg 1 and Leg 2 are 2 sides of the triangle, respectively, and

hypotenuse is the side "opposite" of the 90 degree angle

In this diagram, thus we can write:

r^2 + AB^2 = AO^2

Now putting the information we know, we can solve for r:

r^2 + 9^2 = (10.5)^2

r^2 + 81 = 110.25

r^2 = 110.25 - 81

r^2 = 29.25

r = Sqrt(29.25)

r = 5.41

Rounded to nearest tenth, it is 5.4, 2nd answer chice is right.

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jamal completed his math homework in 2/3 of an hour and his reading homework in 3/5 of an hour. How much time did it take him to
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Step-by-step explanation: We have to add how much time it took him to do all of his homework. To do this, we have to add 2/3 and 3/5.

First, we have to find a LCM between 3 and 5.

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Next we find out what number to multiply 3 by to get 15, we multiply it by 5.

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Calcula y comprueba las ecuaciones: plisss lo necesito alguien me puede ayudar a) 2X = 6 b) 10 + Z = 20 c) P + 9 = 11 d) 3X + 8
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Answer:

a) x=3

b) z=10

c) P= 2

d) X=7

e) U=1

Step-by-step explanation:

Resolver una ecuación consiste en hallar los valores de la variable que hacen cierta la igualdad.

a) 2x= 6

El coeficiente es el número junto a la variable. En este caso, el coeficiente es 2. Para eliminar este número en la expresión 2x, debido a que la variable x esta multiplicada por 2, deberás dividir ambos lados de la ecuación entre 2, debido a que la operación opuesta de la multiplicación es la división.

(2x)÷2=6÷2

x= 3

Comprobar la solución de una ecuación se hace al remplazar la variable en una ecuación con el valor de la solución. La solución debería satisfacer la ecuación cuando se ingresa en esta.

En este caso:

2*3= 6

6=6

b) 10 + z= 20

En este caso se debe sumar o restar la constante que se encuentra acompañando a la variable en ambos lados de la ecuación de manera de aislar el término de la variable. En este caso:

10 - 10 + z= 20 -10

z= 10

Comprobación:

10 + z=20

10 + 10=20

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c) P + 9= 11

P +9 - 9= 11 -9

P=2

Comprobación:

2 + 9= 11

11=11

d) 3X + 8 = 29

En este caso, se suma o resta la constante en ambos lados de la ecuación y luego se elimina el coeficiente de la variable mediante la división o multiplicación. Esto es:

3X + 8 - 8= 29 - 8

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X=7

Comprobación:

3*7 + 8=29

21+8=29

29=29

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2U= 2

2U ÷2= 2÷2

U=1

Comprobación:

2*1 + 8= 10

2 + 8= 10

10=10

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