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3 years ago

What is 2.4 with a bar over it as a fraction..? i need an answer ASAP! ty.

Mathematics

2 answers:

DanielleElmas [232]3 years ago

5 0

Answer:

2 2/5

Step-by-step explanation:

fenix001 [56]3 years ago

5 0

Answer:

2.4/1

Step-by-step explanation:

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40 is 50 percent of which number

FinnZ [79.3K]

Answer:

Step-by-step explanation:

80 divided by 1/2 = 40 therefore 40 is 50% of 80

4 0

3 years ago

What is the distance between the points (−3, 5) and (−3, −6) ?

tresset_1 [31]

The answer would be D, 11 units! :)

8 0

2 years ago

The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.

VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

$\int\limits^2_{-1} {(f(x) - g(x))} \, dx$

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx$

We know that:

$\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}$

$\int\limits^{}_{} {1} \, dx = x$

$\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}$

Then our integral is:

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2} + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3} + (-1) - \frac{(-1)^2}{2} )$

The right side is equal to:

$(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6$

The bounded area is 5 + 5/6 square units.

3 0

2 years ago

Abbey is stacking boxes of candles for store display. The volume of each cube shaped box is 125 cubic inches. If the display has

gtnhenbr [62]

146Answer:146 and your hot

Step-by-step explanation:

4 0

2 years ago

Does anybody know this

Dmitrij [34]

Step-by-step explanation:

The equation used to determine the amount of radioactive material left is

$A = A_0(2)^{-t/HL}$

We are have the following given quantities:

t = 30 seconds

$A_0$ = 50 grams

HL = 6 seconds

Plugging these into our equation we get

$A = (50\:\text{grams})(2)^{-(30s/6s)}$

$\:\:\:\:= (50\:\text{grams})(0.03125)$

$\:\:\:\:= 1.56\:\text{grams}$

6 0

2 years ago