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Kryger [21]
2 years ago
15

Can someone PLEASE help me this is so hardddddd!!!!!

Mathematics
1 answer:
vfiekz [6]2 years ago
8 0

Answer:

hung DJ JJ JD diffused haha song rust iffy Urdu y'all y'all iffy official I

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Without using technology, describe the end behavior of f(x) = −3x38 + 7x3 − 12x + 13.
Vedmedyk [2.9K]
Second option: down on the left, down on the right.

Clearly when x is bigger than one the term -x^38 will dominate the result.

When absolute value of x grows, x^38 becomes a larger positive number and - x^38 a larger negative number. Then both ends of the function tend to negative infinity.
8 0
3 years ago
Read 2 more answers
Help help help thanks
Dmitry [639]

Answer: a=2/5b + 1/5c (I may have answered your question but idk if I did)

Step 1: Add -4a to both sides.

9a−2b+−4a=4a+c+−4a

5a−2b=c

Step 2: Add 2b to both sides.

5a−2b+2b=c+2b

5a=2b+c

Step 3: Divide both sides by 5.

5a/5=2b+c/5

a=2/5b+1/5c

7 0
3 years ago
Let​ T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear​ transformation, and
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

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

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.

6 0
3 years ago
Write two equelivant ratios 11 and 4​
timofeeve [1]

Answer:

The answer is 8 : 22, 12:33 as they are two equelivant ratios 11 and 4

4 0
3 years ago
Mr. Bryon bought a gas can with 4 1/4 gallons of gasoline in it. He used 1/4 of the amount in the can to mow his lawn. How many
alisha [4.7K]
Do 4 1/4 - 1/4 which is easy, just take away the 1/4 and you get 4, so he used 1/4 gallons
6 0
3 years ago
Read 2 more answers
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