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3 years ago

14

At a gas station, the price of a regular gas was $3.80 per gallon on monday. the price increased to $3.85 per gallon on tuesday.

round the price per gallon for each day to the nearest tenth of a dollar.

2 answers:

TEA [102]3 years ago

7 0

Answer:

A round number is less accurate, but easier to use, because you get rid of certain decimals.

There are some rules to do this:

If we want to round a number to the nearest tenth, we have to look what's the unit after it, if that unit is more than 5, we round the tenth to the upper number. For example, 10.66 would be rounded to 10.7.
If we want to round a number, and the next digit is lower that 5, we just maintain the same digit. For example, 10.44 would be rounded to 10.4
However, there's a specific case, if we want to round a number to the tenth, and the following digit is 5, we could round up or round down the tenth digit. If the digit we want to round is pair, then we maintain the same digit, if it's odd, we can round to the upper digit. For example, 10.65 would be rounded to 10.6, and 10.75 would be rounded to 10.8

Therefore, according to these rules, the round number would be $3.8 and $3.8

notsponge [240]3 years ago

4 0

Monday's gas, $3.80 rounded to the nearest tenth of a dollar would be $3.80 (its already there)
Tuesday' gas, $3.85 rounded to the nearest tenth of a dollar would be $3.90
Hope this helps!! and if you ever need anything else just ask! :)

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Every prime number greater than 10 has a digit in the ones place that is include in wich set of numbers A- 1,3,7,9 B-1,3,5,9 D-1

marta [7]

That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:

-- 0:
   A number greater than 10 with a 0 in the units place is a multiple of
   either 5 or 10, so it's not a prime number.

-- 1:
    A number greater than 10 with a 1 in the units place could be
    a prime (11, 31 etc.) but it doesn't have to be (21, 51).

-- 2:
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-- 3:
   A number greater than 10 with a 3 in the units place could be
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-- 4:
   A number greater than 10 with a 4 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 5:
   A number greater than 10 with a 5 in the units place is a multiple
   of either 5 or 10, so it's not a prime number.

-- 6:
   A number greater than 10 with a 6 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 7:
   A number greater than 10 with a 7 in the units place could be
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-- 8:
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   number, and has 2 as a factor, so it's not a prime number.

-- 9:
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   a prime (19, 29 etc.) but it doesn't have to be (39, 69) .

So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.

It CAN't have a 0, 2, 4, 5, 6, or 8 .

The only choice that includes all of the possibilities is 'A' .

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KonstantinChe [14]

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Step-by-step explanation:

The equation of a circle in the center-radius form is:

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Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

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Rearranging the equation:

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Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

<u>For the first parenthesis:</u>

$x^{2}+6x+b^{2}$

We can rewrite this as:

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Hence in this case $b=3$ and $b^{2}=9$ :

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We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

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Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

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ZanzabumX [31]

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Step-by-step explanation:

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2 hours=120 minutes

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