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3 years ago

HELP PLEASE AND THANK YOU 1JAAB

Mathematics

1 answer:

Dafna1 [17]3 years ago

4 0

Answer:

$\bold{Q1}\ side\ BD\\\\\bold{Q2}\ \angle5\ \leftrightarrow\ \angle6\\\\\bold{Q3}\ \triangle TAP\\\\\bold{Q4}\ \triangle APT$

Step-by-step explanation:

For Q1 and Q2

$\text{If}\ AB\cong AC,\ BD\cong CD,\ RSTU\ \text{is a parallelogram (rectangle)},\ \text{then}\\\\\angle 1\ \leftrightarrow\ \angle2\\\angle3\ \leftrightarrow\ \angle4\\\angle C\leftrightarrow\ \angle B\\\angle 5\ \leftrightarrow\ \angle6\\\angle7\ \leftrightarrow\ \angle 8\\\angle R\ \leftrightarrow\ \angle T\\\angle S\ \leftrightarrow\ \angle U$

$\overline{AB}\ \leftrightarrow\ \overline{AC}\\\overline{BD}\ \leftrightarrow\ \overline{CD}\\\\\overline{RS}\ \leftrightarrow\ \overline{TU}\\\overline{ST}\ \leftrightarrow\ \overline{UR}$

For Q3 and Q4

$O\ \leftrightarrow\ T\\B\ \leftrightarrow\ A\\R\ \leftrightarrow\ P\\\\E\ \leftrightarrow\ X\\F\ \leftrightarrow\ W\\D\ \leftrightarrow\ T$

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2 years ago

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

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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;

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