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4 years ago

If the orientation of two shapes is not the same, which transformation(s) should you use to show they are congruent? Why?

Mathematics

2 answers:

barxatty [35]4 years ago

7 0

Answer:

Reflection and Rotation

Step-by-step explanation:

If the orientation of two shapes is not the same, first check which directions they point. Changing the orientation of a shape to map it onto another is only possible through either reflection or rotation. So, choose a suitable transformation—either reflection or rotation—to map one shape onto the other. Translations alone would not show the congruence of shapes with different orientations.

Aleksandr-060686 [28]4 years ago

4 0

Answer:

The answer is below

Step-by-step explanation:

Transformation s the movement of a point from its initial location to a new location. If an object is transformed, the all its points are also transformed. Types of transformation are rotation, reflection, translation and dilation.

Two shapes are congruet if they have the same shape and size, two objects are congruent if they can be moved without changing their shape or size. Congruent transformations are transformation that create congruent objects. The three types are:

Translation: sliding an object either up, down, left or right.
Reflection: Flipping an object over a line
Rotation: turning an object about a point

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garik1379 [7]

Fyi area=pir^2 not 3r2
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3 years ago

12p + 15 - 13 - 12p in simplest form

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Answer:

Step-by-step explanation:

First, lets order it by like terms.

12p - 12p + 15 -13

Then, simplify

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3 years ago

At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given

Nostrana [21]

Answer:

a. The initial amount was 10 mg.

b. The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d. The time until only 1 mg of the drug remains in the body is 11.6 hours.

Step-by-step explanation:

You know that at time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by :

$A=10*(0.82)^{t}$

a. The initial quantity occurs when time t is the initial t, that is, t is equal to 0. Then:

$A=10*(0.82)^{t}=10*(0.82)^{0}=10*1\\$

A=10

The initial amount was 10 mg.

b. Considering that an exponential growth is determined by:

$A=A0*(1-r)^{t}$ , where A is the amount after a certain number of a certain time, Ao is the initial amount, r is the rate and t is the time, so the percentage of the drug that leaves the body each hour is :

1-r=0.82

Solving:

1-r -1= 0.82 -1

-r= -0.18

r= 0.18

The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is when t = 6:

$A=10*(0.82)^{6}$

Solving:

A= 3.04 mg

The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d. The time that passes until only 1 mg of the drug remains in the body is calculated taking into account that A = 1 mg:

$1=10*(0.82)^{t}$

Solving:

$0.1=(0.82)^{t}$

㏒ 0.1= t*㏒ 0.82

㏒ 0.1 ÷ ㏒ 0.82= t

11.6 hours= t

The time until only 1 mg of the drug remains in the body is 11.6 hours.

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3 years ago

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RoseWind [281]

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Step-by-step explanation:

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3 years ago

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marishachu [46]

Answer:

thanks for free point Mark me as brain list

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