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Troyanec [42]
3 years ago
12

A microscope shows you an image of an object that is 20 times the objects actual size. Therefore, the scale factor of the enlarg

ement is 20. An insect has a body length of 10 millimeters. What is the body length of the insect under the microscope?
Mathematics
2 answers:
madam [21]3 years ago
4 0
We just have to multiply 20 by 10mm, which is 200mm. 100mm is equal to one centimeter, so the body length of the insect under the microscope is two centimeters.
AVprozaik [17]3 years ago
3 0

Answer:

200 millimeters


your welcome XD



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Which expressions are equivalent to the expression 7(5 + m) - 8?
Akimi4 [234]

Answer:

<u>2</u><u>7</u><u> </u><u>+</u><u> </u><u>7</u><u>m</u>

Step-by-step explanation:

7(5 + m) - 8

open the bracket:

= 35 + 7m - 8 \\  = 27 + 7m

3 0
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
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
What is 2/3 + 3/4 I don’t understand it at all
victus00 [196]

Answer:

6/12 or half

Step-by-step explanation:

this would be the answer because the lowest common denominator of 4 and 3 is 12 so you'd multiply them you'd get 12 then multiply 2 and 3 and get six therfore the answer is 6/12 or 1/2

6 0
3 years ago
Read 2 more answers
Which of the following is a polynomial with roots: − square root of 5 , square root of 5 , and −3?
tresset_1 [31]
Roots: - √5 , √5, and - 3

=> these are factors of the polynomial: (x + √5), (x - √5), and (x + 3).

Multiply those three factors:

(x + √5) (x - √5) ( x + 3) = [x^2 - 5] ( x + 3 ) = x^2 + 3x^2 - 5x - 15

Therefore the polynomial x^2 + 3x^2 - 5x - 15 is a polynomial with the given roots.

Answer: option B. x^3 + 3x^2 - 5x - 15
4 0
3 years ago
The Bernards are trying to determine their babysitter's rates but are to embarrassed to just ask her. They were charged $20 for
Alina [70]

Answer:

A = 5.5t + 9

Step-by-step explanation:

We assume that the relationship between t and A is linear.

We have two points on a line: (2, 20) and (4, 31).

We now find the equation of the line.

A = mt + b

m = (31 - 20)/(4 - 2) = 11/2 = 5.5

20 = 5.5(2) + b

20 = 11 + b

b = 9

A = 5.5t + 9

3 0
3 years ago
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