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2 years ago

14

The pink trapezoid is dilated by a factor of 3. The resulting image is shown in blue. What is the ratio of the perimeters of the

two trapezoids? (small:large)
A) 1:2

B) 1:3

C) 1:4

D) 1:6

2 answers:

Reil [10]2 years ago

8 0

B) 1:3

The ratio is 1 to 3

laila [671]2 years ago

4 0

Solution: 1:3. The two trapezoids are similar. Therefore, the ratio of the perimeters is the same as the ratio of the side lengths.

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The ratio of the geometric mean and arithmetic mean of two numbers is 3:5, find the ratio of the smaller number to the larger nu

IgorC [24]

Answer:

$\frac{1}{9}$

Step-by-step explanation:

Let the numbers be x,y, where x>y

The geometric mean is

$\sqrt{xy}$

The Arithmetic mean is

$\frac{x + y}{2}$

The ratio of the geometric mean and arithmetic mean of two numbers is 3:5.

$\frac{ \sqrt{xy} }{ \frac{x + y}{2} } = \frac{3}{5}$

We can write the equation;

$\sqrt{xy} = 3$

or

$xy = 9 - - - (2)$

l

and

$\frac{x + y}{2} = 5$

or

$x + y = 10 - - - (2)$

Make y the subject in equation 2

$y = 10 - x - - - (3)$

Put equation 3 in 1

$x(10 - x) = 9$

$10x - {x}^{2} = 9$

${x}^{2} - 10x + 9 = 0$

$(x - 9)(x - 1) = 0$

$x =1 \: or \: 9$

When x=1, y=10-1=9

When x=9, y=10-9=1

Therefore x=9, and y=1

The ratio of the smaller number to the larger number is

$\frac{1}{9}$

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2 years ago

Jamar's car used 1/40 of a gallon to travel 3/8 of a mile. How many miles can the car go on one gallon of gas?

dybincka [34]

Answer:

15 miles.

Step-by-step explanation:

3/8 x 40 = 120/8

now divide,

120/8 = 15

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2 years ago

Write the standard equation of the circle with center (1,2) and radius of 2

MrMuchimi

Answer:

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Step-by-step explanation:

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2 years ago

Correct and best answer gets brainliest.

Nuetrik [128]

Im pretty sure the answer is c) 136

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3 years ago

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Match the equations with their solutions over the interval [0, 2π].

nignag [31]

I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs are:

cos(x)tan(x) – ½ = 0 → π/6 and 5π/6

cos(π/6)tan(π/6) – ½ = 0

cos(5π/6)tan(5π/6) – ½ = 0

sec(x)cot(x) + 2 = 0 → 7π/6 and 11π/6

sec(7π/6)cot(7π/6) + 2 = 0

sec(11π/6)cot(11π/6) + 2 = 0

sin(x)cot(x) + 1/sqrt2 = 0 → 3π/4 and 5π/4

sin(3π/4)cot(3π/4) + 1/sqrt2 = 0

sin(5π/4)cot(5π/4) + 1/sqrt2 = 0

csc(x)tan(x) – 2 = 0 → π/3 and 5π/3

csc(π/3)tan(π/3) – 2 = 0

csc(5π/3)tan(5π/3) – 2 = 0

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3 years ago

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