The values of a and b are 4 & 6
Answer:
Karen had 45 m and m's candy.
Step-by-step explanation:
Let the number of m and m's candy be 'x'.
Now given:
Karen gave an equal amount of m and m's to herself and four friends.
So we can say that;
Number of people m and m's candy distributed equally = 5
Also Given:
Each person receives m and m's equivalent to the largest one digit number.
Now we know that;
Largest one digit number is 9.
So we can say that;
Each person receives m and m's = 9
We need to find number of m and m's Karen have.
Solution:
So we can say that;
Total number of m and m's Karen have is equal to Number of people m and m's candy distributed equally multiplied by number of m and m's can each person receives.
framing in equation form we get;
Total number of m and m's Karen had = 
Hence Karen had 45 m and m's candy.
Answer:
False
Step-by-step explanation:
To solve this problem, plug in the value for Y, 9, into the inequality.
9 + 3 < 12
Now, solve the inequality. Evaluating 9 + 3, the equation becomes:
12 < 12
12 is not less than 12 (12 = 12), therefore the answer is false.
Answer:
I believe the answer is D :)
Answer:
1) False
2) False
3) True
4) False
Step-by-step explanation:
1) Flase, {v1,v2,v3, ..., vp} is a base for H when they span H and also they are linearly independent.
2) False. A single nonzero vector is linearly independent , not dependent. There is not null linear combination that gives 0 as a result involving that vector.
3) True, if the columns werent linearly independent, we could triangulate the matrix and obtain 0, so the matrix wouldnt be invertible. This means that the columns should be linearly independent for the matrix to be invertible and as a consecuence, they will spam a subspace of R^n of dimension n, which means that they will spam all R^n and therefore, they form a basis of R^n.
4) False. A basis is a spanning set that is as small as possible. Larger spanning sets will have extra elements apart from those who can form a base toguether. Those elements will make the set linearly dependent.
