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3 years ago

(1.9x10^-3)(2x10^4) Show your work

Mathematics

1 answer:

Helga [31]3 years ago

7 0

The correct answer is

(1.9 * 10∧-3) * (2 * 10∧4) = 1.9 * 2 * 10∧-3 * 10∧4 = 3.8 * 10∧(-3+4) =3.8 * 10∧1=

= 3.8 * 10 = 38

Good luck!!!

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A rental car company charges $53.75 per day to rent a car and $0.12 for every mile driven. Jaya wants to rent a car, knowing tha

Aliun [14]

Answer:

Jaya can afford to rent a car 4 days while staying within her budget.

Step-by-step explanation:

The inequality would have to indicate that the cost of renting a car has to be less than or equal to $230. The cost to rent a car is equal to the cost per day for the number of days plus the price per mile for the number of miles, which is:

53.75x+0.12y≤230, where:

x is the number of days the car is rented

y is the number of miles driven

As the statement says that she plans to drive 125 miles, you can replace "y" with this value and solve for x:

53.75x+0.12(125)≤230

53.75x+15≤230

53.75x≤230-15

53.75x≤215

x≤215/53.75

x≤4

According to this, the answer is that Jaya can afford to rent a car 4 days while staying within her budget.

5 0

3 years ago

20 POINTS !! IF YOU ANSWER THE QUESTIONS !

tatuchka [14]

12 the answer is 12 I know it

6 0

3 years ago

Sherry’s pay rate is $18.75 per hour; last week she worked 50 hours. Calculate her gross pay for the week.

saveliy_v [14]

Answer:

937.5

Step-by-step explanation:

$18.75 - 1 hour

$x. - 50hours

cross multiply the 18.75 with 50.

5 0

3 years ago

3x − 1/2(8x − 2) = 4 step 1: 3x − 4x 1 = 4 step 2: x 1 = 4 step 3: x 1 − 1 = 4 − 1 step 4: x = 3 which step was the error made a

Mrrafil [7]

3x-1/2(8x-2)=4
3x-4x+1=4
-x=3
X=-3

8 0

4 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

4 years ago