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Alex Ar [27]
3 years ago
9

First determine the missing angle for triangle 1 and triangle 2. Then apply the angle-angle criterion to determine if the two tr

iangles are similar. Choose all that are similar. Triangle 1 – ∠1 = 50°, ∠2 = 30° Triangle 2 – ∠2 = 20°, ∠3 = 100° Triangle 1 – ∠1 = 60°, ∠2 = 20° Triangle 2 – ∠1 = 40°, ∠3 = 100° Triangle 1 – ∠1 = 25°, ∠2 = 115° Triangle 2 – ∠1 = 25°, ∠3 = 40° Triangle 1 – ∠1 = 5°, ∠2 = 15° Triangle 2 – ∠2 = 15°, ∠3 = 160°
Mathematics
1 answer:
barxatty [35]3 years ago
7 0

Answer:

  • Triangle 1 – ∠1 = 25°, ∠2 = 115° Triangle 2 – ∠1 = 25°, ∠3 = 40°
  • Triangle 1 – ∠1 = 5°, ∠2 = 15° Triangle 2 – ∠2 = 15°, ∠3 = 160°

Step-by-step explanation:

You can save yourself some trouble if you realize that one of the angles in one pair must match one of the angles in the other pair.

This observation eliminates the first two choices.

__

Going further, the triangles will be similar if the dissimilar angles together with one of the similar angles totals 180°.

__

Triangle 1 – ∠1 = 50°, ∠2 = 30°

Triangle 2 – ∠2 = 20°, ∠3 = 100°  . . . . . no angles match

__

Triangle 1 – ∠1 = 60°, ∠2 = 20°

Triangle 2 – ∠1 = 40°, ∠3 = 100°  . . . . . no angles match

__

Triangle 1 – ∠1 = 25°, ∠2 = 115°

Triangle 2 – ∠1 = 25°, ∠3 = 40° . . . . 25° angles match; 25+40+115 = 180

These triangles are similar.

__

Triangle 1 – ∠1 = 5°, ∠2 = 15°

Triangle 2 – ∠2 = 15°, ∠3 = 160° . . . . 15° angles match; 15+5+160 = 180

These triangles are similar.

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In 2073 BS father's age was 7 times his daughter devyani's age. In 2081 BS father's age will be 3 times only his daughter's age.
Anna007 [38]

Answer:

2069.

Step-by-step explanation:

Let Devyani's age in 2073= x years and y be her fathers age  Then:

y = 7x

- and in 8 years time ( 2081) we have:

y + 8 = 3(x + 8)

y = 3x + 24 - 8

y = 3x  + 16.     Eliminating y from the 2 equations:

7x = 3x + 16

4x = 16

x = 4.

So in 2073 she was 4 years old

Therefore she was born in 2069.

4 0
3 years ago
Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.
lesya [120]

Answer:

0.1994 is the required probability.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851

P(sample of 20 boxers is more than 167 pounds)

P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)

= 1 - P(z \leq 0.8438)

Calculation the value from standard normal z table, we have,  

P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0
3 years ago
Please help me out!!!!!!!!
qaws [65]

Answer:

x = - 6

Step-by-step explanation:

Given that y varies directly as x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = - 9 when x = 3, thus

k = \frac{y}{x} = \frac{-9}{3} = - 3

y = - 3x ← equation of variation

When y = 18, then

18 = - 3x ( divide both sides by - 3 )

x = - 6

5 0
3 years ago
I need help with this. screenshot attached
CaHeK987 [17]
Answer: 7/5

rise/run
= (7-0)/(8-3) = 7/5
7 0
3 years ago
Bill can buy jags, jigs and jogs for $\$1$, $\$2$ and $\$7$ each, respectively. What is the largest number of jogs he can purcha
bekas [8.4K]

Answer:

6 jogs

Step-by-step explanation:

To spend exactly $50

Jags = $1

Jigs = $2

Jogs = $7

If he must buy atleast 1 of each and still spend exactly $50

Hence, the largest number of jogs he can buy can be modeled as :

First obtain the multiples of 7 up to 50

Multiple of 7 up to 50

7 : 7, 14, 21, 28, 35, 42, 49

If $49 is spent on jigs, then he won't be able to purchase jigs.

Hence, the largest amount he can spend on jogs is $42

Thus, he can buy $42 / $7 = 6 jogs

The $8 left can be spent on jigs and jags

7 0
3 years ago
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