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Phoenix [80]
3 years ago
13

Please help me on this question

Mathematics
1 answer:
IRINA_888 [86]3 years ago
5 0
The most precise is 301.4m
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Jill had a twenty meter rope. she cut it into eight equal pieces. how long was each piece?
NARA [144]
Its 2.5 20/8 equals this plz mark brainlest
3 0
4 years ago
Solve the equation 2p/3-12=-2
-BARSIC- [3]
Hey there! 

\frac{2p}{3} - 12 = -2

Firstly, we are going add 12 on each of the sides that we're working with. like ↓ 

p - 12 + 12 \\ \\ = -2 + 12

This gives us \frac{2}{3}p = 10 (if you are wondering how we got the out come of 10 it is because I added -2 + 12 = 10

Now multiply \frac{3}{2} on each of your sides 

\frac{3}{2} ( \frac{2}{3}p)  \\ \\  =  \frac{3}{2} (10)

Cancel the first set and you will find the value of p

p = 15

Good luck on your assignment and enjoy your day 

~LoveYourselfFirst:)
5 0
3 years ago
Ana is a scuba diver who has completed 5 dives for a total depth of 230 feet. Choose an equation using integers to represent the
galben [10]

Answer:

Step-by-step explanation:

-230/5=-46

7 0
3 years ago
The length and width of a rectangle are measured as 30cm and 24cm, respectively, with an error in measured of
larisa [96]

Answer:

The maximum error in the calculated area of rectangle is 5.4 cm².

Step-by-step explanation:

Given : The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measured of  at most 0.1 cm in each.

To find : Use differentials to estimate the maximum error in the calculated area of rectangle ?

Solution :

The area of the rectangle is A=l\times w

The derivative of the area is equal to the partial derivative of area w.r.t. length times the change in length plus the partial derivative of area w.r.t. width times the change in width.

i.e. dA=\frac{\partial A}{\partial L}\Delta L+\frac{\partial A}{\partial W}\Delta W

Here, \frac{\partial A}{\partial L}=30,\ \frac{\partial A}{\partial W}=24 ,\ \Delta L=0.1,\ \Delta W=0.1

Substitute the values,

dA=(30)(0.1)+(24)(0.1)

dA=3+2.4

dA=5.4

Therefore, the maximum error in the calculated area of rectangle is 5.4 cm².

7 0
3 years ago
Helppppppppppppppppppppppppppppppppppppp
sineoko [7]

Answer:

Option C is correct.........

4 0
3 years ago
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