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aksik [14]
1 year ago
10

Which expression can be used to convert 100 USD to Japanese yen?

Mathematics
2 answers:
katen-ka-za [31]1 year ago
5 0

Since,

1 USD = 113.83 Japanese Yen

100 USD = 100 x 113.83 Japanese Yen

100 USD = 11,383.00 Japanese Yen

maxonik [38]1 year ago
3 0

Answer:

a. 100usd{99.30487yen/1usd}

Step-by-step explanation:

you welcome

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A law firm charges a client according to the function f(x)=4(x+2) , where x represents the number of hours spent on the client’s
Zigmanuir [339]

Answer:

The graph in the attached figure

Step-by-step explanation:

Let

x ---> the number of hours spent on the client’s case each week

f(x) ---->  the total charge that week in hundreds of dollars

we have

f(x)=4(x+2)

distribute

f(x)=4x+8

This is the equation of the line in slope intercept form

where

The slope or unit rate is

m=4

The y-intercept or initial value is

b=8

To graph the line we need minimum two points

Find the x-intercept

For f(x)=0

0=4x+8\\4x=-8\\x=-2

The x-intercept is the point (-2,0)

so

we have

the points (-2,0) and (0,8)

To graph the line, plot the intercepts, connect them and join the points

The graph in the attached figure

Remember that the value of x and the value of f(x) cannot be a negative number

3 0
2 years ago
What number comes next 16, 06, 68, 88,
inessss [21]

So the given series is "16, 06, 68, 88, __"

Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).

From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.

Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.

4 0
2 years ago
You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the
Makovka662 [10]

Answer:

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

Step-by-step explanation:

a)  How much will you have at the middle of the first year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

To determine:

Total amount = A = ?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

substituting the values

A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}

A=300\cdot \frac{2.06}{2}

A=\frac{618}{2}

A=309 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

To determine:

Total amount = A = ?

so using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

so substituting the values

A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}

A=300\cdot \frac{2.06^2}{2^2}

A=318.27 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

3 0
1 year ago
During the first seconds after takeoff, a rocket traveled 208 kilometers in 50 minutes at a constant rate suppose a penny is dro
andrezito [222]

the penny travels at a faster rate by 56.4 kilometers

penny = 153 for 30 min

penny = 306 for 1 hour

rocket = 208 in 50 min

208/5 = 41.6 *6 = 249.6 in 1 hour

penny is faster by 56.4 kilometers

6 0
2 years ago
A barrel of crude oil contains about 5.61 cubic feet of oil. How many barrels of oil contained in 1 mile (5280 feet) of a pipeli
8_murik_8 [283]
1)volume of the  pipeline

The pipeline is a cylinder, therefore;
Volume (cylinder)=πr²h
r=radius
h=height of the cylinder

diameter=6 in*(1 ft / 12 in)=0.5 ft
raius=diameter / 2=0.5 ft / 2=0.25 ft.
height=5280 ft

Volume (pipeline)=π(0.25 ft)²(5280 ft)=330π ft³≈1036.73 ft³.


2) we calculate the number of barrel
1 mile of oil in this pipeline is 330π ft³ of oil.

1 barrel of crude------------------5.61 ft³
x----------------------------------330π ft³

x=(1 barrel*330π ft³) / 5.61 ft³=184.8 barrels

3) we calculate the price.

1 barrel---------------$100
184.8 barrels----------  x

x=(184.8 barrels * $100) / 1 barrel=$18,480

Solution: ≈$18,480
6 0
2 years ago
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