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3 years ago

What does simple interest mean

Mathematics

1 answer:

notka56 [123]3 years ago

3 0

It is the quick and easy method of calculating the interest charge on a loan.

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Select the correct answer. Which expression is equivalent to this polynomial? 16x² +4

lana66690 [7]

Answer:

(4x+2)(4x-2)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A contaminant is leaking into a lake at a rate of R(t) = 1700e^0.06t gal/h. Enzymes that neutralize the contaminant have been ad

olasank [31]

Answer:

16,460 gallons

Step-by-step explanation:

This is a differential equation problem, we have a constant flow of contaminant into the lake, but also we know that only a fraction of that quantity of contaminant remains because of the enzymes. For that reason, the differential equation of contaminant's flow into the lake would be:

$\frac{dQ}{dt} =1700exp(0.06t)*exp(-0.32t)\\\frac{dQ}{dt} =1700exp(-0.26t)\\$

Then, we have to integrate in order to find the equation for Q(t), as the quantity of contaminant in the lake, in function of time.

$\int\limits^0_t {dQ}=\int\limits^0_t {1700exp(-0.26t)dt}\\Q(t)=\frac{1700}{-0.26} exp(-0.26t)+C \\$

Now, we use the given conditions to replace them in the equation, in order to solve for $C$

$t_{0} =0\\Q_{0}=10,000\\Q_{0}=-6538exp(-0.26*0)+C\\C=10,000+6538=16538$

Then, we reorganize the equation and we replace t for 17 hours, in order to determine the quantity of contaminant at that time:

$Q_{t} =-6538exp(-0.26t)+16538\\Q_{17} =-6538exp(-0.26*17)+16538\\Q_{17} =16460 gallons$

3 0

3 years ago

What is the answer to this math problem 3/10+2/5

Dominik [7]

Multiply 2/5 by 2 to get 4/10

4/10 + 3/10 = 7/10

4 0

4 years ago

Read 2 more answers

5x+1/7-2x-6/4>-4 solve the linear equation

8_murik_8 [283]

5x+1/7-2x-6/4>-4 equals x>-37/42

8 0

4 years ago

Let f(x) = (x − 1)2, g(x) = e−2x, and h(x) = 1 + ln(1 − 2x). (a) Find the linearizations of f, g, and h at a = 0. What do you no

sweet [91]

Answer:

Lf(x) = Lg(x) = Lh(x) = 1 - 2x

value of the functions and their derivative are the same at x = 0

Step-by-step explanation:

Given :

f(x) = (x − 1)^2,

g(x) = e^−2x ,

h(x) = 1 + ln(1 − 2x).

a) Determine Linearization of f, g and h at a = 0

L(x) = f (a) + f'(a) (x-a) ( linearization of f at a )

for f(x) = (x − 1)^2

f'(x ) = 2( x - 1 )

at x = 0

f' = -2

hence the Linearization at a = 0

Lf (x) = f(0) + f'(0) ( x - 0 )

Lf (x) = 1 -2 ( x - 0 ) = 1 - 2x

For g(x) = e^−2x

g'(x) = -2e^-2x

at x = 0

g(0) = 1

g'(0) = -2e^0 = -2

hence linearization at a = 0

Lg(x) = g ( 0 ) + g' (0) (x - 0 )

Lg(x) = 1 - 2x

For h(x) = 1 + ln(1 − 2x).

h'(x) = -2 / ( 1 - 2x )

at x = 0

h(0) = 1

h'(0) = -2

hence linearization at a = 0

Lh(x) = h(0) + h'(0) (x-0)

= 1 - 2x

Observation and reason

The Linearization is the same in every function i.e. Lf(x) = Lg(x) = Lh(x) this is because the value of the functions and their derivative are the same at x = 0

8 0

3 years ago