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3 years ago

The simple interest earned on a deposit of $4,000 at 5% for 120 days.

Mathematics

2 answers:

liraira [26]3 years ago

7 0

Answer:

Option B $\$66.67$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$I=Prt$

where

P is the Principal amount of money to be invested

r is the rate of interest in decimal form

t is Number of Time Periods in years

in this problem

we assume that

$1\ month=30\ days$

$t=120\ days=(1/3)\ years\\ P=\$4,000\\r=0.05$

substitute in the formula above

$I=\$4,000*0.05*(1/3)=\$66.67$

fgiga [73]3 years ago

5 0

B. I=PRT is the simple interest formula. 120 days is 4 months, or 1/3 of a year.
I=(4,000)(0.05)(0.33333)
I=$66.67

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Integrate 1 - x / x(x2 + 1) d x by partial fractions.

solniwko [45]

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:- by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$ ......(1)

step 2:-

solving on both sides

$1-x=A(x^{2} +1)+(Bx+C)x......(2)$

substitute x =0 value in equation (2)

1=A(1)+0

A=1

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

0 = 1+B

B=-1

comparing x co-efficient on both sides (in equation 2)

-1 = C

step 3:-

substitute A,B,C values in equation (1)

now

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$

by using integration formulas

i) by using $\int\limits \frac{1}{x} d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\$ .....(b)

$\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C$ .....(c)

step 4:-

by using above integration formulas (a,b,and c)

we get answer is

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

6 0

3 years ago

After painting the fence steve and janet realize that their house also needs painting. But steve will have to do the painting hi

gladu [14]

Answer:

Therefore, Steve will paint house for 20.6 days.

Step-by-step explanation:

We know that Steve determines that it would take them 9 days to paint the house together (if they were both healthy) and that it would take Janet (when healthy) 16 days to paint the house alone.

We conclude that in 1 day Janet painted 1/16 of the house. In nine days, Janet painted 9/16 of house.

So, 7/16 of house paint Steve in nine days.

We have the following proportion:

$\frac{7}{16}:9=1:x\\\\\frac{7}{16}x=9\\\\x=9\cdot \frac{16}{7}\\\\x=20.6$

Therefore, Steve will paint house for 20.6 days.

4 0

3 years ago

PLEASE HELP WITH GRAPHING QUESTION ASAP!!

bulgar [2K]

x-int: -0.5; y-int: 1

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3 years ago

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A system of linear equations has been graphed in the diagram. determine a reasonable solution for the system of equations.

Grace [21]

Answer:

(1,-1)

Step-by-step explanation:

The solution to a system of equation means the x and y values if we were to solve both equations (of lines) simultaneously.

That is also known as the intersecting points.

The top-right part is known as the 1st Quadrant in a coordinate system.

Top-left is 2nd Quadrant.

Bottom left is 3rd Quadrant.

Bottom right is 4th Quadrant.

The values of x and y (positive or negative) of each quadrant is shown below:

Quadrant 1: x +ve, y +ve
Quadrant 2: x -ve, y +ve
Quadrant 3: x -ve, y -ve
Quadrant 4: x +ve, y -ve

The intersection point (solution) of the 2 lines shown is in the 4th quadrant, which has x values positive and y values negative.

From the answer choices, we see that 3rd answer choice goes with this -- (1,-1), so this is the correct choice.

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3 years ago

Solve 3y^{2} - (y + 2) (y - 2)

hram777 [196]

Answer:

3y^2 - (y + 2) (y - 2) = 0

<=> 3y^2 - (y^2 - 4) = 0

<=> 2y^2 + 4 =0

<=>y^2 + 2 = 0

=> Because y^2 is always equal or larger than 0, there is no real solution.

Hope this helps!

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3 years ago

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