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Paul [167]
2 years ago
14

1c.What is his new texting speed if he practices for 3 months

Mathematics
2 answers:
dexar [7]2 years ago
8 0
His new texting speed will be 3.5 characters per second this is because of the texting rate increasing per month is 0.5, just multiply that by 3 because there is 3 months than add that to 2 seconds
Hunter-Best [27]2 years ago
6 0

Answer:

3\frac{1}{2} character per second.

Step-by-step explanation:

James current texting speed = 2 characters per second

Increase in texting speed with practice = \frac{1}{2} a character per second per month.

If he practices for x months then the expression that will represent his typing speed = 2 + \frac{1}{2}\times x

If x = 3 months then his typing speed after 3 months will be

Speed = 2+\frac{3}{2}

           = \frac{4+3}{2}

           = \frac{7}{2}

           = 3\frac{1}{2} character per second.

Therefore, after 3 months of practice his typing speed will be 3\frac{1}{2} character per second.

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4 0
3 years ago
Which expression is equivalent to (a^-3 b/a^-5 b3)^-3? Assume a=0, b=0.
prohojiy [21]
Use:\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(ab)^n=a^nb^n

\left(\dfrac{a^{-3}b}{a^{-5}b^3}\right)^{-3}=\left(a^{-3-(-5)}b^{1-3}\right)^{-3}=\left(a^{-3+5}b^{-2}\right)^{-3}=\left(a^2b^{-2}\right)^{-3}\\\\=(a^2)^{-3}(b^{-2})^{-3}=a^{2(-3)}b^{-2(-3)}=a^{-6}b^6=\dfrac{b^6}{a^6}
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3 years ago
Solve the in quality x plus 1<7
RUDIKE [14]

Answer:

x<6

Step-by-step explanation:

x+1 < 7

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7 0
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Let E be the event where the sum of two rolled dice is greater than or equal to 7. List the outcomes in Ec.
zhenek [66]

Answer:

(1, 1)  (1, 2)  (1, 3) (1,4)

(1,5)  (2,1)   (2,2) (2,3)

(2,4)  (3,1)  (3,2)   (3,3)

(4,1)   (4,2) (5,1)

Step-by-step explanation:

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Answer:

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