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3 years ago

Jose can type 30 words in 6 minutes. What is his rate in words per minute?

Mathematics

2 answers:

Minchanka [31]3 years ago

7 0

Answer:

5 words a minute

Step-by-step explanation:

30 words in 6 minutes= 30 words/6 minutes= 5 words a minute.

Check:

5 words/minute * 6 minutes= 30 words/minute

kompoz [17]3 years ago

5 0

Answer: 180 wpm

Step-by-step explanation: 30x6=180

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zalisa [80]

To write this in standard form, you need to eliminate the fraction in the coefficient of variable x. You can do this by multiplying 8 to the two sides of the equation:

8(y) = 8[(-5/8)x + 3]

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Transpose the variable x to the other side:

5x + 8y = 24

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3 years ago

A store sells pool floats.

Andrei [34K]

Step-by-step explanation:

the original retail price was 120% of the manufacturer price (100% manufacturer price + 20% markup).

now, the whole retail price is reduced by 12%.

that means we have to subtract 12% from the 120%.

but careful, we cannot just do 120-12=108.

that is simply because these 120% are now for this calculation the new basic amount and therefore 100%.

we actually need to calculate and subtract 12% of 120 to get the new percentage over the manufacturer price.

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so, the new retail price is now only 5.6% over the manufacturer price.

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2 years ago

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PilotLPTM [1.2K]

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2 years ago

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Akimi4 [234]

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3 years ago

Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4

lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

f(x) = { x^2+3x+4 , x < 3

f(x) = { x^2+3x+4 , x≥3

Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:

$f(x) = x^2+3x+4$

- The area "A1" bounded by f(x) is given as:

$A1 = \int\limits^a_b {f(x)} \, dx$

Where, The interval of the function { -3 , 3 ] = [ a , b ]:

$A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2$

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

$f(x) = 4x+10$

- The area "A2" bounded by f(x) is given as:

$A2 = \int\limits^a_b {f(x)} \, dx$

Where, The interval of the function { 3 , 4 ] = [ a , b ]:

$A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2$

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

A = A1 + A2

A = 42 + 26

A = 68 unit^2

8 0

3 years ago