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horrorfan [7]
3 years ago
7

Hey please help, thank you!!

Mathematics
1 answer:
marin [14]3 years ago
5 0

Answer:

D

Step-by-step explanation:

So we have the equation:

A=\frac{1}{2}(12)(3+7)

And we want to find out which of the answer choices is <em>not</em> equivalent.

Thus, let's go through each of the choices.

A)

We have:

\frac{12(3+7)}{2}

This is correct.

Our original equation can be written as:

A=\frac{1}{2}(\frac{12(3+7)}{1})

And if we multiply straight across:

A=\frac{12(3+7)}{2}

We'll get choice A. So choice A is correct.

B)

We have:

6(3+7)

This is also correct.

From our original equation:

A=\frac{1}{2}(12)\cdot(3+7)

If we multiply the first term together:

A=6(3+7)

We'll get choice B. So, choice B is correct.

C)

We have:

0.5(12)(10)

This is correct as well.

If we reduce 1/2 to decimal form and add the operations within the parentheses in the original equation, we will get:

A=\frac{1}{2}(12)(3+7)\\A=0.5(12)(10)

So, C is also correct.

D)

We have:

\frac{12}{2}\times\frac{10}{2}

Note that this is not correct.

This attempted to use the distributive property. However, the distribute property does <em>not work</em> for multiplication. For instance:

3(2+1)\text{ indeed equals } 3(2)+3(1)

However:

3(2\times1)\text{ does not equal} 3(2)\times3(1)

So, choice D is not equivalent.

So choice D is our answer :)

If we evaluate it, we will get 30, in contrast to the 60 of the others.

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Klio2033 [76]

Answer:

\{T(v_1), T(v_2), T(v_3)\} is linearly dependent set.

Step-by-step explanation:

Given:  \{v_1,v_2,v_3\} is a linearly dependent set in set of real numbers R

To show: the set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

Solution:

If \{v_1,v_2,v_3,...,v_n\} is a set of linearly dependent vectors then there exists atleast one k_i:i=1,2,3,...,n such that k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0

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A linear transformation T: U→V satisfies the following properties:

1. T(u_1+u_2)=T(u_1)+T(u_2)

2. T(au)=aT(u)

Here, u,u_1,u_2∈ U

As T is a linear transformation,

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\

As \{v_1,v_2,v_3\} is a linearly dependent set,

k_1v_1+k_2v_2+k_3v_3=0 for some k_i\neq 0:i=1,2,3

So, for some k_i\neq 0:i=1,2,3

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

Therefore, set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

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Answer:

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Step-by-step explanation:

Angle is ending at the point = (-3, -4)

Signs of both x and y are negative therefore it is in the third quadrant.

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Answer:

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