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MakcuM [25]
2 years ago
10

Write an expression to find how many times Jason will blink in 60 minutes if he continues to blink 19 times a minute.

Mathematics
1 answer:
expeople1 [14]2 years ago
7 0

Answer:

19*60

Step-by-step explanation:

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Help me with this math question
nasty-shy [4]

Answer:

\huge \boxed{K=20}\checkmark

B. 20

Step-by-step explanation:

<em>First, you divide by 5 from both sides of equation.</em>

<em>\displaystyle \frac{5k}{5}=\frac{100}{5}</em>

<em>Simplify, to find the answer.</em>

<em>\displaystyle 100\div5=20</em>

<em>\huge \boxed{K=20}, which is our answer.</em>

6 0
3 years ago
Read 2 more answers
Find the slope between (-3,1) and (-17,2)
lapo4ka [179]

Answer:

-1/14

Step-by-step explanation:

7 0
3 years ago
Pls answer with real answers only thank you
aev [14]
Answer: 32.35 cm^2

Step by step:

Find the area of the rectangle first.

A= L • W
A= 11 • 4.2
A= 46.2 cm^2

Then find the area of the circle. The formula is A= pi (r)^2. The diameter of the circle is 4.2 cm because looking at the width of the rectangle it fits into the circle as well.

Half of the diameter is 2.1 cm which is the radius.

A= pi (r)^2
A= pi (2.1)^2
A= pi (4.41)
A= 13.85 cm^2

Then you would subtract 13.85 from 46.2 to find the shaded portion.

Hope this helps :))

4 0
2 years ago
The number of students in an school building that have the flu after t days is given by the function
gulaghasi [49]

Step-by-step explanation:

p(0) =  \frac{800}{1 + 49 {e}^{ - 0.2 \times 0} }  =  \frac{800}{1 + 49}  =  \frac{800}{50}  = 16

200 =  \frac{800}{1 + 49 {e}^{ - 0.2t} }  \\ 200(1 + 49 {e}^{ - 0.2t} ) = 800 \\ 200 + 9800 {e}^{ - 0.2t}  = 800 \\ 9800 {e}^{ - 0.2t}  = 600 \\  {e}^{ - 0.2t}  =  \frac{3}{49}  \\  ln( {e}^{ - 0.2t} )  =  ln( \frac{3}{49} )  \\ -  0.2t =  - 2.793 \\ t = 13.96 = 14

6 0
3 years ago
The question is in the picture​
trasher [3.6K]

Answer:

should also = 35* if I'm not mistaken

Step-by-step explanation:

if this is wrong I'm very sorry. if no one else helps u with in the next 10 mins I suggest using socratic, it helps me

8 0
3 years ago
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