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WINSTONCH [101]
3 years ago
6

look at the figure shown below which step should be used to prove that point a is equidistant from points c and b

Mathematics
1 answer:
JulijaS [17]3 years ago
3 0
D
is certainly wrong. You could extend the length of AD as far as you want and the two triangles (ABD and ACD) would still be congruent.

C
is wrong as well. The triangles might be similar, but they are more. They are congruent.

B
You don't have to prove that. It is given on the way the diagram is marked.

A
A is your answer. The two triangles are congruent by SAS 
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Answer:

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Step-by-step explanation:

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The answer of the larger circle = 254.34
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The altitude of a triangle forms a perpendicular line.<br><br> False<br><br> True
Ksivusya [100]
True is the answer for this question
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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
Matrix A has 2 rows and 3 columns whereas matrix B has3 rows and 2 column
sergeinik [125]

Step-by-step explanation:

for any matrix multiplication number of columns of first matrix should be equal to number of rows of second matrix.

For AB

A has 3 columns and B has 3 rows so it matches . Hence can be multiplied.

For BA

B has 2 columns and A has 2 rows it also matches so can be multiplied

8 0
2 years ago
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