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arlik [135]
3 years ago
6

A pitchers arm rotates at a speed of 7 degrees per millisecond

Mathematics
1 answer:
Fantom [35]3 years ago
7 0

Answer:

To solve this problem, all we have to do is to use a conversion factor to convert millisecond into seconds. We know that there are 1000 milliseconds in a second, therefore:

pitcher’s arm rotation = 7 degrees / millisecond * (1000 milliseconds / second)

pitcher’s arm rotation = 7,000 degrees / second

Step-by-step explanation:

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X = 3y - 6 2x - 4y = 8 Solve the system of equations using substitution.
Brut [27]
Hello

x = 3y - 6
2x - 4y = 8
We need to start by solving 3y - 6 for x
Substitute 3y -6 for x in 2x
2x - 4y = 8
2(3y - 6) - 4y = 8
2y - 12 = 8
2y = 8 + 12
2y = 20
Divide both sides by 2
2y/2 = 20/2
y = 10

Now, substitute 10 for y in x= 3y
x = 3y - 6
x = 3(10) - 6
x = 30 - 6
x= 24

Answer : x = 24 and y=10

I hope this help!
7 0
3 years ago
Read 2 more answers
Harrison has 30 cds in his music collection. If 40% of the cds are country and 30% are pop, how many are other types of music
Pavel [41]
Well you have to find out how many cd is 40% and 30% 
30% = 30x.3=9
40% = 30x.4=16
now add 16 and 9
16+9=25 now subtract 25 from 30 and see that you have 5 cd of a different type of music to find that as a % divide 5 by 35 to get .14 I rounded it to the nearest hundreth now move the decimal over 2 spots to get 14%
8 0
3 years ago
Read 2 more answers
Solve -7x^2+x+9=-6x quadratic formula
myrzilka [38]

Answer:

-7x^2+x+9=-6x

-7x^2+x+9+6x=-6x+6x

-7x^2+7x+9=0

quadratic\: formula

x_{1,\:2}=\frac{-7\pm \sqrt{7^2-4\left(-7\right)\cdot \:9}}{2\left(-7\right)}

\sqrt{-7^{2} -4(-7)\times9} =\sqrt{301}

x_{1,\:2}=\frac{-7\pm \sqrt{301}}{2\left(-7\right)}

\frac{-7+\sqrt{301}}{2\left(-7\right)}

=\frac{-7+\sqrt{301}}{-2\cdot \:7}

=\frac{-7+\sqrt{301}}{-14}

\frac{-7-\sqrt{301}}{2\left(-7\right)}

=\frac{-7-\sqrt{301}}{-2\cdot \:7}

=\frac{-7-\sqrt{301}}{-14}

answer=\frac{7+\sqrt{301}}{14}

OAmalOHopeO

7 0
3 years ago
Bearing a to b is 280 what is bearing b to A
jeka94

Answer:

please see photo attached for detailed analysis.

5 0
2 years ago
A 6m ladder leans against a wall. The bottom of the ladder is 1.3m from the wall at time =0sec and slides away from the wall at
icang [17]

Answer:

- 0.100

Step-by-step explanation:

Length of the ladder,  H = 6 m

Distance at the bottom from the wall, B = 1.3 m

Let the distance of top of the ladder from the bottom at the wall is P

Thus,

from  Pythagoras theorem,

B² + P² = H²    .

or

B² + P² = 6²          ..............(1)      [Since length of the ladder remains constant]

at B = 1.3 m

1.3² + P² = 6²

or

P² = 36 - 1.69

or

P² = 34.31

or

P = 5.857

Now,

differentiating (1)

2B(\frac{dB}{dt})+2P(\frac{dP}{dt})=0

at t = 2 seconds

change in B = 0.3 × 2= 0.6 ft

Thus,

at 2 seconds

B = 1.3 + 0.6 = 1.9 m

therefore,

1.9² + P² = 6²

or

P = 5.69 m

on substituting the given values,

2(1.9)(0.3) + 2(5.69) × (\frac{dP}{dt}) = 0

or

1.14 + 11.38 × (\frac{dP}{dt}) = 0

or

11.38 × (\frac{dP}{dt}) = - 1.14

or

(\frac{dP}{dt}) = - 0.100

here, negative sign means that the velocity is in downward direction as upward is positive

5 0
3 years ago
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