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3 years ago

Multiply 12.57 by 4.11. Round your answer to the nearest hundredth.

1 answer:

4 0

When you are going to multiply 12.57 by 4.11, the result would be 51.6627.

To round off the result to the nearest hundredth, it would become 51.66.

According to the rules of rounding off a number, if the number after the rounded off digit is more than or equal to 5, you have to add 1 to the rounded off digit. But if it is less than 5, then the rounded off digit would be still the same. In this case, the rounded off digit which is 6 would still be the same since the number after that is 2 which is less than 5.
So, the answer is letter B which is 51.66

You might be interested in

Rounding each number to the nearest tenth

seraphim [82]

Answer: (17.310 would round up to 18) (7.14 would round down to just 17) So your new equation will be 18 + 17 =35

Step-by-step explanation:

4 0

4 years ago

1. Your distance from lightning varies directly with the time it takes you to hear thunder

raketka [301]

Answer: d = 1/5t

Step-by-step explanation:

$\\$ Since it is a direct variation , Let d represent the distance and t represent the time ,then

$\\$ d ∝ t

$\\$ introducing the proportionality constant, we have

$\\$ d = kt

$\\$ Substituting t = 10 and d = 2 , we have

$\\$ 2 = 10k

$\\$ k = 1/5

$\\$ substitute k = 1/5 into the original equation , we have

$\\$ d = 1/5t

4 0

3 years ago

Evaluate the expression when b=3 and x= -2. -b + 8x

jolli1 [7]

b=3 and x= -2.

-b + 8x

-3 + 8(-2)

-3 + -16

-19

have a great day :)

7 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

4 years ago

A bakery sells one cupcake for $2.75 or a dozen cupcakes for $25.80. If you

wlad13 [49]

Answer:

D $0.60

Step-by-step explanation:

Since a dozen is equal to 12, we will first multiply $2.75 by 12, and we get $33. Now subtract $33 by $25.80 dollars, which is $7.2. Now we know that we saved $7.2 out of a dozen cupcake's, but we have to find how much we saved on each cupcake. So lets divide $7.2 by 12 and we get $0.60.  Therefore, you saved $0.60 on each cupcake.

5 0

3 years ago