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Bogdan [553]
3 years ago
13

Please help if you know the answer please

Mathematics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

1 3/7 or 10/7

Step-by-step explanation:

2 is also known as 14/7. 14/7 - 4/7 is equal to 10/7 or 1 3/7.

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An airplane is ascending at a constant rate of 15 feet per second what is the change in altitude during 10 minutes of flight
fenix001 [56]

Answer:

9000 ft

Step-by-step explanation:

Given: A airplane is ascending at a constant rate of 15 feet per second,

What we need to know: The change in altitude during 10 minutes of flight,

60 seconds = 1 minute

60 × 10

= 600 seconds

600 × 15

= 9000 ft

Therefore there was a 9000 ft change in altitude during the 10 minutes of the flight.

6 0
2 years ago
Y = 5x + 7<br> 2x + y = -5<br> Are the equations parallel,<br> perpendicular, or neither?
Nezavi [6.7K]

Lines are neither perpendicular nor parallel.

Step-by-step explanation:

Step 1:

In equation 1, Y = 5X + 7

In equation 2,  Y = -2X -5

Step 2:

Slope Intercept form Y =mX +c

Where m = Slope

Slope of line 1 = 5

Slope of line 2 = -2

m1  is not equal to m2. Hence it is not parallel.

m1  × m2 is not equal to -1. Hence they are not perpendicular.

3 0
3 years ago
What is 60 x 60 - 12 =? fp ifykyk
salantis [7]

Answer:

3588

Step-by-step explanation:

60*60=3600

3600-12

6 0
3 years ago
Read 2 more answers
Read the excerpt from Heart of a Samurai.
Mazyrski [523]

Answer: D. the snail’s path

Step-by-step explanation: because The author uses the most precise words to describe the snail's path.

This excerpt has been taken from Heart of a Samurai, a historical novel written by Margi Preus in 2010. Precise words such as 'shiny', 'ribbon' and 'unfurling' have been used in the second sentence to describe the path of the snail, which is a small animal that has a wet body and moves very slowly. In other words, the author describes the way the path looks by establishing a comparison between the path and a shiny ribbon.

Hope this help

4 0
3 years ago
Read 2 more answers
Simplify sin q + cosq cotq.
IRINA_888 [86]
\sin q+\cos q\cot q=\sin q+\cos q\cdot\dfrac{\cos q}{\sin q}=\sin q+\dfrac{\cos^2q}{\sin q}\\\\=\dfrac{sin^2q}{\sin q}+\dfrac{\cos^2q}{\sin q}=\dfrac{\sin^2q+\cos^2q}{\sin q}=\dfrac{1}{\sin q}=\csc q\\\\Used:\\\cot q=\dfrac{\cos q}{\sin q}\\\\\sin^2q+\cos^2q=1\\\\\csc q=\dfrac{1}{\sin q}
3 0
3 years ago
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