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3 years ago

car for $50,000. If the car depreciates 10% per year, use an exponential function to find the value of the car in 7 years.

Mathematics

1 answer:

Ymorist [56]3 years ago

4 0

50000x0.90 to the power of 7

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A calculator screen shows a number in scientific notation as 4.18E‒8. Write this number in standard form.

Citrus2011 [14]

Answer:

.0000000418

Step-by-step explanation:

So you will start from the decimal and move 8 places to the left.

E-8 is the same as 10^-8 so you will move to the left

If it was E8 you would move to the right 8 places.

6 0

3 years ago

After being rearranged and simplified, which of the following equations could

masha68 [24]

Answer:

C and D

Step-by-step explanation:

The quadratic formula is

x= (-b±√b²-4ac)/2a

The formula uses the numerical coefficients in the quadratic equation.

The general quadratic equation is ax²+bx+c where a, b and c are the numerical coefficients

So, lets try and see;

$5x+4=3x^4-2\\\\=3x^4-5x-2-4\\=3x^4-5x-6\\a=3,b=-5,c=-6$

But due to the fact that in this equation you have x⁴, the equation is not a quadratic equation thus can not be solved using this formula

$-x^2+4x+7=-x^2-9\\\\\\=-x^2+x^2+4x+7+9\\=4x+16$

$9x+3x^2=14+x-1\\\\\\=3x^2+9x-x-14+1\\\\=3x^2+8x-13\\\\\\a=3,b=8,c=-13\\$

$2x^2+x^2+x=30\\\\\\=3x^2+x-30\\\\\\a=3,b=1,c=-30$

From the checking above, the equations will be C and D

7 0

2 years ago

Read 2 more answers

200 labourers need 50 days to construct a road of 4 km long. How many labourers are needed to construct 8 km long in 80 days

patriot [66]

Answer:

250

Step-by-step explanation:

Number of labourers * Number of days worked = Total number of working days

Total number of working days to construct a 4km long road: $200*50=10000$

Total mumber of working days to construct a 8km long road: $10000*2=20000$

Labourers: 20000 ÷ 80 = 250.

250 labourers are required.

7 0

3 years ago

Brian invests £1300 into his bank account.

raketka [301]

Answer:

£1690

Step-by-step explanation:

Amount invested by Brian = £1300

rate of simple interest = 10%

To find money Brian will have after three years

He will have amount invested in bank and interest earned in three years from that amount.

Simple interest for any principal amount p is given by

SI = P*R * T /100

where SI is simple interest earned

T is time period for which simple interest is earned

R is rate of interest

Substituting value of P , R and T we have

SI = 1300*10* 3 /100 = 390

Therefor interest earned will be £390

Total money with Brian after three years = principal amount invested + interest earned in 3 years

= £1300 + £390 = £1690

3 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago