Zuma was earning $9.40 per hour when she got a 6% increase in pay. What was her rate of pay after the increase

1 answer:

6 0

$9.96 To do the math take your starting value and multiply 9.40 by 1.06 (1 in the 1's place because you are just adding 6%) (0.6 in the hundredths place due to the 6/100% or 06/100%)

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A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it

Troyanec [42]

Answer:

W = 15/4 ft . ib

Step-by-step explanation:

Force = 10ib

According to hooked law, f(x) = kx

x = 4inches = 4/12 ft

x= 1/3ft

f(x) = 1/3k

10 = 1/3k

k = 30 ib/ft

f(x) = 30x

Workdone = integral of f(x) with its limit

6 inches = 6/12 ft

= 1/2ft

W = integral(1/2 to 0) of 30x

W = 15x^2(1/2 to 0)

W = 15(1/2)^2 - 15(0)^2

W = 15(1/4) - 0

W = 15/4 ft. Ib

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3 years ago

for a field trip 12 students rode in cars and the rest filled six buses how many students were in each bus if 342 students were

kolezko [41]

Answer:

Step-by-step explanation:

we have 342 students, subtract 12 because we need only kids riding the bus, now we have 330 kids, we divide the 330 kids between 6 buses 330/6, our answer is 55.

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3 years ago

A blogger has 400 subscribers to her blog in January. The number of subscribers has grown by a factor of 1.5 every month since t

mr Goodwill [35]

Answer:

Since a blogger has 400 subscribers to her blog in January, and the number of subscribers has grown by a factor of 1.5 every month since the, to write a sequence to represent the number of subscribers in the 3 months that followed, the following reasoning has to be made:

January: 400 subscribers

February: 400 x 1.5 = 600 subscribers

March: 600 x 1.5 = 900 subscribers

April: 900 x 1.5 = 1,350 subscribers

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3 years ago

If my saving x$ grows 10% every year how much will I have in: 1 year 2 year 5 year

mote1985 [20]

Answer:

$1.1x, 1.21x, 1.61051x$

Step-by-step explanation:

If you saving grows $10 \%$ every year, then your saving is $1.1\\$ times your saving from last year. Therefore, after one year, you will have $1.1x\\$ , then after $2$ years, you will have $(1.1)^2 \cdot x=1.21x$ , then after $5$ years, you will have $(1.1)^5 \cdot x = 1.61051x$ .