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AlekseyPX
3 years ago
9

Mitchell and his friends went to the candy store, where they can buy candy by the pound. Mitchell bought 1.2 pounds of candy. Hi

s friends bought 2.3 pounds and 1.8 pounds. About how many pounds of candy did they buy altogether?
rounding
Mathematics
2 answers:
sergiy2304 [10]3 years ago
3 0

Answer:

5.3 pounds

Step-by-step explanation:

1.2 plus 1.8 is 3 pounds

then you add 3 pounds plus 2.3 lbs and you get

5.3 pounds

Keith_Richards [23]3 years ago
3 0

Answer:

5.3

Step-by-step explanation:

If you add how much Mitchell bought (1.2) and how much his friends bought (2.3), (1.8) therefore the answer is 5.0 or rounded is just 5

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1 year ago
Please simplify 10{N + 3).
lana [24]

Answer:

Step-by-step explanation:

I must assume that you meant 10(N + 3).  Let's not mix { with ).

10(N + 3) is as simple as this expression gets.  You could, however, re-write it as 10N + 30.

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3 years ago
Gene paid a deposit on a leased car. The deposit earns 2.8 percent simple annual interest. At the end of the year, the interest
Lapatulllka [165]

Answer:

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6 0
3 years ago
Read 2 more answers
A card is randomly selected from a standard
shepuryov [24]

Answer:

3 / 13

Step-by-step explanation:

Hello!

There are 52 cards in a deck with half being red and half being black. The question ask for when a black card is drawn so we divide the total by 2

52/ 2 = 26

There are 6 black face cards in a deck so to get the probability one will be drawn we put how many there are over the total amount

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8 0
3 years ago
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Evaluate the following integral using trigonometric substitution.
wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function \int\limits {\sqrt{a^{2} -x^{2} } } \, dx, using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as asin \theta i.e x = a sin\theta.

All integrals in the form \int\limits {\sqrt{a^{2} -x^{2} } } \, dx are always evaluated using the substitute given where 'a' is any constant.

From the given integral, \int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx where a = 7 in this case.

The substitute will therefore be   x = 7 sin\theta

2.) Given x = 7 sin\theta

\frac{dx}{d \theta} = 7cos \theta

cross multiplying

dx = 7cos\theta d\theta

3.) Rewriting the given integral using the substiution will result into;

\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta  } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)}   } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)}   } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)}   }}} \, 7cos\theta d\theta\\

= \int\limits343 cos^{2}  \theta \, d\theta

8 0
3 years ago
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