Evaluate <br> 2(x-4)+3x-x^2 for x=3

Pls help me and pls explain how u got answer

nignag [31]

Answer:

q+4 + d-3 + n ; first

Step-by-step explanation:

first :q quarter d dimes n nickels

then: q+4 quarter d dimes n nickels

then: q+4 quarter d-3 dimes n nickels

so you have q+4 + d-3 + n

first is the only option that has division the other involve subtraction

6 0

3 years ago

Read 2 more answers

648 ÷ 8 tell how you solved pls i need help

stira [4]

Answer:

The answer is 81

Step-by-step explanation:

All you have to do is create your long division chart with 648 inside and 8 on the outside 8 can not go into 6 but it can go into 64 so 8 goes on top of 4 you subtract 64 minus 64 becayse 8 times 8 is 64. Lastly you bring down the other 8 and 8 goes into 8 once so you put 1 on top. And so thats how you get 81.

8 0

3 years ago

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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

the number of students in the four sixth-grade classs at northside school are 26, 19,34 and 21. Use properties to find the total

Aleks04 [339]

In this case all you do is add all the numbers, answer is 100

4 0

4 years ago

Plz Help

GenaCL600 [577]

Begin by factoring 2 out of 2x^2 - 2x - 12 equals 0:

2(x^2 - x - 6) = 0

2(x - 3)(x + 2) = 0. 2 is never zero, but x-3 and x+2 can each be set = to 0:

This results in x = 3 and x = -2. The equation is true for these two x-values.

6 0

4 years ago

Evaluate 2(x-4)+3x-x^2 for x=3