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3 years ago

How many faces does a right rectangular prism have if it has 8 vertices and 12 edges?

Mathematics

2 answers:

Ganezh [65]3 years ago

5 0

Answer:

Step-by-step explanation:

I think

Rzqust [24]3 years ago

5 0

Vas happenin!!

A rectangular prism has 4 faces

So A is correct

Hope this helps

-Zayn Malik

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Graphic image of the kite jklm after a reflection across the line Y= 1

olchik [2.2K]

Answer: see image

<u>Step-by-step explanation:</u>

Draw the line y = 1. Count how many units away the point is from that line. Move that point the same number of units in the opposite direction to create the reflection.

J is 2 units above y = 1. The new point is 2 units below y = 1. → (3, -1)

K is 3 units above y = 1. The new point is 3 units below y = 1. → (6, -2)

L is 5 units above y = 1. The new point is 5 units below y = 1. → (3, -4)

M is 3 units above y = 1. The new point is 3 units below y = 1. → (0, -2)

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All real numbers / all x
if you're asking for range as well then its -1<x<1

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Work out m and c for the line: y -4x= − 1

Lelu [443]

Step-by-step explanation:

$y = 4x - 1 \\ m = 4 \\ c = - 1$

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Common denominator of 5 6 7

Yanka [14]

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5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ... 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ... 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...

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3 years ago

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

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2 years ago