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Aleksandr [31]
3 years ago
14

2 1/2 divided by 4. can you please answer i need help

Mathematics
2 answers:
marissa [1.9K]3 years ago
8 0

Answer:

5/8

Step-by-step explanation:


poizon [28]3 years ago
6 0
The answer would be 5/8, 62.5% in percentage mode, and 0.625 in decimal.

Convert 2 1/2 to a
improper fraction

Simplify 2 * 2 to 4

Simplify 4 + 1 to 5

Simplify

Simplify 2 * 4 to 8

Now fuse it together and it would be 5/8.
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Nicole simplifies 2^a*2^5/2^b to 2^(a-b+5). Which two properties of exponents did she use while simplifying the expression?
emmainna [20.7K]

Answer:

D.

Step-by-step explanation:

2^a*2^5/2^b = 2^(a+5)/2^b. using power of a product property.

then

2^(a+5)/2^b= 2^(a+5)*2^-b. Using Negative Exponent property.

Again we can use power of a product property and we get 2^(a+5)*2^-b=2^(a-b+5)

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3 years ago
Differentiate y=xe^(-x/2).
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We will use the product rule:
d/dx( f(x) · g(x) ) = f´( x ) g( x ) + g´( x ) f( x )    ( and after that the chain rule )
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=e^{- \frac{x}{2} } + x * e^{- \frac{x}{2} } *(- \frac{1}{2})=  \\ e^{- \frac{x}{2} } - \frac{x}{2} * e^{- \frac{x}{2} } = e ^{- \frac{x}{2} } (1- \frac{x}{2} )
4 0
3 years ago
Make a tree diagram for two coins that are tossed. Find the theoretical probability that at least one coin is heads. Express the
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8 0
3 years ago
A 4 meter tall bear is walking in the woods and sees honey flavoured chips strapped to a flashlight hanging from a tree 7 meters
otez555 [7]

Answer:

The length of the shadow is decreasing 8 metres per second.

Step-by-step explanation:

In the attached figure, we draw the situation of the bear approaching the tree.

We can use the Tales to make some calculations.

The proportion between the height of the bear (h) and his shadow (S) is equal to the proportion between the height of the flashlight (H) and the sum of the distance of the bear (d) and the bear shadow.

This can be written as:

\frac{h}{S} =\frac{H}{S+d}

If we clear S we have:

S*H=S*h+d*h\\\\(H-h)*S=d*h\\\\S=(\frac{h}{H-h}) d

We know that the distance is a function of time and it is reduced, as the bear approaches the tree, as a rate of 6 m/s.

We can express then the rate of variation of the shadow S as:

\Delta S=S_2-S_1=(\frac{h}{H-h}) (d_2-d_1)=(\frac{h}{H-h}) \Delta d\\\\\\\Delta d=-6\\\\H=7\\\\h=4\\\\\\\Delta S=(\frac{4}{7-4}) *(-6)=\frac{4}{3} *(-6)=-8

The length of the shadow is decreasing 8 metres per second.

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Step-by-step explanation:

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