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Flauer [41]
1 year ago
6

In the diagram of circle o, what is the measure of ABC? 27° 54° 108° 120°

Mathematics
1 answer:
Vlad1618 [11]1 year ago
5 0

<u>Given</u>:

Given that a circle O with two tangents BA and BC.

The major arc AC is 234°

The minor arc AC is 126°

We need to determine the measure of ∠ABC

<u>Measure of ∠ABC:</u>

We know the property that, "if a tangent and a secant, two tangents or two secants intersect in the interior of the circle, then the measure of angle formed is one half the difference of the measures of the intercepted arcs."

Hence, applying the above property, we have;

\angle ABC=\frac{1}{2}( major \widehat{AC} - minor \widehat{AC})

Substituting the values, we get;

\angle ABC=\frac{1}{2}( 234^{\circ} -126^{\circ})

\angle ABC=\frac{1}{2}( 108^{\circ})

\angle ABC=54^{\circ}

Thus, the measure of ∠ABC is 54°

Hence, Option b is the correct answer.

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Two cells are viewed and measured under a microscope. The approximate diameter
nikklg [1K]

Answer:

option D. 4.7 x 10-4 m

Step-by-step explanation:

<u><em>The correct question is</em></u>

• cell P: 5.0 x 10-4 meters

• cell Q: 3.0 x 10-5 meters

we have            

The diameter of cell P is  5*10^{-4}\ m

The diameter of cell Q is  3*10^{-5}\ m

Rewrite the diameter of cell P as    

5*10^{-4}\ m=5*10^{-4}(\frac{10}{10})=50*10^{-5}\ m

Find the difference

50*10^{-5}\ m-3*10^{-5}\ m=47*10^{-5}\ m

Rewrite the difference as

47*10^{-5}\ m=47*10^{-5}}(\frac{10}{10})=4.7*10^{-4}\ m

7 0
2 years ago
In triangle RST, U is the midpoint of RS, V is the midpoint of ST, and W is the midpoint of TR. Use the triangle diagram to answ
Zolol [24]

(1)

we are given

U is the midpoint of RS

and we have

RU=12

so, we can use formula

RS=2RU

we can plug values

RS=2\times 12

RS=24............Answer

(2)

V is the midpoint of ST

so, we get

SV=VT

now, we can plug values

2x=11

divide both sides by 2

x=5.5.........Answer

(3)

now, we can find y

W is the midpoint of TR

so, we get

RW=WT

we can plug value

15.9=3y

divide both sides by 3

y=5.3

(4)

we can see that

triangles URW and RST are similar

so, their sides ratios must be equal

so, we get

\frac{RW}{RT} =\frac{UW}{ST}

we can plug values

\frac{15.9}{2\times 15.9} =\frac{UW}{2VT}

\frac{15.9}{2\times 15.9} =\frac{UW}{2\times 11}

UW=22\times \frac{15.9}{2\times 15.9}

UW=11...........Answer

(5)

we can see that

triangles SUV and RST are similar

so, their sides ratios must be equal

so, we get

\frac{SU}{SR} =\frac{UV}{RT}

now, we can plug values

\frac{12}{2\times 12} =\frac{UV}{2\times 15.9}

UV=2\times 15.9\times \frac{12}{2\times 12}

UV=15.9.............Answer

4 0
1 year ago
La fuerza necesaria para evitar que un auto derrape en una curva varía inversamente al radio de la curva y conjuntamente con el
Vika [28.1K]

Answer:

768 libras de fuerza

Step-by-step explanation:

Tenemos que encontrar la ecuación que los relacione.

F = Fuerza necesaria para evitar que el automóvil patine

r = radio de la curva

w = peso del coche

s = velocidad de los coches

En la pregunta se nos dice:

La fuerza requerida para evitar que un automóvil patine alrededor de una curva varía inversamente con el radio de la curva.

F ∝ 1 / r

Y luego con el peso del auto

F ∝ w

Y el cuadrado de la velocidad del coche

F ∝ s²

Combinando las tres variaciones juntas,

F ∝ 1 / r ∝ w ∝ s²

k = constante de proporcionalidad, por tanto:

F = k × w × s² / r

F = kws² / r

Paso 1

Encuentra k

En la pregunta, se nos dice:

Suponga que 400 libras de fuerza evitan que un automóvil de 1600 libras patine alrededor de una curva con un radio de 800 si viaja a 50 mph.

F = 400 libras

w = 1600 libras

r = 800

s = 50 mph

Tenga en cuenta que desde el

F = kws² / r

400 = k × 1600 × 50² / 800

400 = k × 5000

k = 400/5000

k = 2/25

Paso 2

¿Cuánta fuerza evitaría que el mismo automóvil patinara en una curva con un radio de 600 si viaja a 60 mph?

F = ?? libras

w = ya que es el mismo carro = 1600 libras

r = 600

s = 60 mph

F = kws² / r

k = 2/25

F = 2/25 × 1600 × 60² / 600

F = 768 libras

Por lo tanto, la cantidad de fuerza que evitaría que el mismo automóvil patine en una curva con un radio de 600 si viaja a 60 mph es de 768 libras.

7 0
2 years ago
Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take
MrRissso [65]

Answer:

mean (μ) = 4.25

Step-by-step explanation:

Let p = probability of a defective computer components = \frac{68}{400} = 0.17

let q = probability of a non-defective computer components = \frac{332}{400} = 0.83

Given random sample n = 25

we will find mean value in binomial distribution

The mean of binomial distribution = np

here 'n' is sample size and 'p' is defective components

mean (μ) = 25 X 0.17 = 4.25

<u>Conclusion</u>:-

mean (μ) =  4.25

6 0
1 year ago
the driving distance between Manchester and London is 195 miles. Faris intends to travel from Manchester to London by coach. The
Dahasolnce [82]

Answer:

7.24pm

Step-by-step explanation:

196+50=3.9h

0.9h=0.9x60mnt=54mnt

3.9h=3h 54mnt

3:30pm+3h 54mnt

3:30pm+3h 54mnt=7.2pm

3 0
1 year ago
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