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snow_lady [41]
3 years ago
11

A nurse worked from 8:15 am until 11:30 am and 12:30 pm until 4:15 pm. Estimate the total number of hours worked.

Mathematics
1 answer:
grin007 [14]3 years ago
7 0
The total numbers of hours worked would be 20
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Select the correct answer from each drop-down menu.
ella [17]

Answer:

A non-equilateral rhombus.

Step-by-step explanation:

We can solve this graphically.

We start with square:

ABCD

with:

A = (11, - 7)

B = (9, - 4)

C = (11, - 1)

D = (13, - 4)

Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:

AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13

BC =  √( (11 - 9)^2 + (-1 -(-4))^2) = √13

CD =  √( (11 - 13)^2 + (-1 -(-4))^2)  = √13

DA =  √( (11 - 13)^2 + (-7 -(-4))^2) = √13

And we change C by C' = (11, 1)

In the image you can see the 5 points and the figure that they make:

The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.

5 0
2 years ago
A spinner with 9 equal sections is numbered 1 through 9. The probability of spinning a 1 or a 9 is 2/9 .
USPshnik [31]
The answer would be 7/9. If the spinner has 9 sections and you want to know the probability of NOT landing on a 1 or a 9, you would subtract 2 from 9 since there are 2 numbers and 9 total.

6 0
3 years ago
Read 2 more answers
Which point could represent 5 3 ?
Fittoniya [83]

Answer:

the answer is c

Step-by-step explanation:

i just finished the lesson

7 0
3 years ago
Read 2 more answers
Fractions that are greater than 3/8
BaLLatris [955]

Answer:

do it yourself looser

Step-by-step explanation:

3 0
3 years ago
Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere is V = 4 3 πr3 and its surface
Scrat [10]

Answer:

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.

Step-by-step explanation:

The area of a sphere is given by the following formula:

A = 4\pi r^{2}

In which A is the area, measured in cm², and r is the radius, measured in cm.

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min.

This means that \frac{dr}{dt} = 40

Determine the rate of change in surface area when r = 20 cm.

This is \frac{dA}{dt} when r = 20. So

A = 4\pi r^{2}

Applying implicit differentiation.

We have two variables, A and r, so:

\frac{dA}{dt} = 8r\pi \frac{dr}{dt}

\frac{dA}{dt} = 8*20\pi*40

\frac{dA}{dt} = 20106.19

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.

6 0
3 years ago
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