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Strike441 [17]
2 years ago
11

A triangle with vertices at A(0,0), B(0,4), and C(6,0) is dilated to yield a triangle with vertices at A(0,0), (B(0,10), and C (

C(15,0). The origin is the center of dilation. What is the scale factor of the dilation
Mathematics
1 answer:
dezoksy [38]2 years ago
4 0

The required scale factor for the dialed triangle is given by SF= 6.28

A triangle with vertices at A(0,0), B(0,4), and C(6,0) is dilated to yield a triangle with vertices at A(0,0), (B(0,10), and C (C(15,0).

<h3>What is scale factor?</h3>

The scale factor is defined as the ratio of modified change in length to the original length.

Since, the height and base of the triangle is y coordinate of B and x coordinate of c is given by 4, 6
Area of triangle =1/2(base x height)
                         = 1/2 (4 x 6)
                         =   4.8
Similarly, for dilated triangle
Area of dilated triangle =  30

Now, scale factor(SF) = area of dilated triangle \ area of triangle
SF = 30/4.8
SF = 6.25

Thus, the required scale factor is SF = 6.25

 

Learn more about line Scale factor:

brainly.com/question/22312172

#SPJ1

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TiliK225 [7]
450/25=18; 18x10=180. therefore the answer is B: 180 inches
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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
Evaluate, when x=2 y=-5 and z=3
gregori [183]

Answer:

-139

Step-by-step explanation:

Evaluate 1/4 (4 x^3 - 2 y - 2 z^3) y^2 - 16 x^2 where x = 2, y = -5 and z = 3:

(4 x^3 - 2 y - 2 z^3)/4 y^2 - 16 x^2 = (4×2^3 - -5×2 - 2×3^3)/4×(-5)^2 - 16×2^2

(4×2^3 - 2 (-5) - 2×3^3)/4×(-5)^2 = ((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4:

((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4 - 16×2^2

(-5)^2 = 25:

((4×2^3 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2^3 = 2×2^2:

((4×2×2^2 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2^2 = 4:

((4×2×4 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2×4 = 8:

((4×8 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

3^3 = 3×3^2:

((4×8 - 2 (-5) - 23×3^2) 25)/4 - 16×2^2

3^2 = 9:

((4×8 - 2 (-5) - 2×3×9) 25)/4 - 16×2^2

3×9 = 27:

((4×8 - 2 (-5) - 227) 25)/4 - 16×2^2

4×8 = 32:

((32 - 2 (-5) - 2×27) 25)/4 - 16×2^2

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((32 + 10 - 2×27) 25)/4 - 16×2^2

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| 3 | 2

+ | 1 | 0

| 4 | 2:

(42 - 54 25)/4 - 16×2^2

42 - 54 = -(54 - 42):

(-(54 - 42) 25)/4 - 16×2^2

| 5 | 4

- | 4 | 2

| 1 | 2:

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(-12)/4 = (4 (-3))/4 = -3:

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2^2 = 4:

-3×25 - 164

-3×25 = -75:

-75 - 16×4

-16×4 = -64:

-64 - 75

-75 - 64 = -(75 + 64):

-(75 + 64)

| 7 | 5

+ | 6 | 4

1 | 3 | 9:

Answer:  -139

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Answer:

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