Bianca wrote the steps for her solution to the equation 2.3 + 8(1.3x – 4.75) = 629.9. She left out the description for the last
step.
What is the missing step of her solution?
A: Add 10.4 to both sides.
B: Add negative 10.4 to both sides.
C: Multiply both sides of the equation by 10.4.
D: Divide both sides of the equation by 10.4.
What is the missing step of her solution?
A: Add 10.4 to both sides.
B: Add negative 10.4 to both sides.
C: Multiply both sides of the equation by 10.4.
D: Divide both sides of the equation by 10.4.
2 answers:
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Answer:
= 78
= 3.5
Step-by-step explanation:
First we need to find .
We can use the equation to solve for
.
We can then change that equation to , since the Commutative Property of Addition says that you can have any addition in any order.
Now, we can solve the equation.
Now that we solved , we can now solve for
.
Since 25 equals , we can solve the equation
.
Here is how you solve it:
Since equals 3.5, which is the simplest form, that is the answer.
Hope this helps, and please mark me brainliest! :)
Answer:
(A) Set A is linearly independent and spans . Set is a basis for
.
Step-by-Step Explanation
<u>Definition (Linear Independence)</u>
A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.
<u>Definition (Span of a Set of Vectors)</u>
The Span of a set of vectors is the set of all linear combinations of the vectors.
<u>Definition (A Basis of a Subspace).</u>
A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.
Given the set of vectors , we are to decide which of the given statements is true:
In Matrix , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column.
has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans
.
Therefore Set A is linearly independent and spans . Thus it is basis for
.