Bonnie is making a dipping sauce. She mixes 150 milliliters of soy sauce with 100 milliliters of vinegar. How much soy sauce doe
1 answer:
Answer:
Bonnie is mixing 1.50 milliliter of soy sauce with 1 milliliter of vinegar
Step-by-step explanation:
Given
Bonnie mixes 150 milliliters of soy sauce with 100 milliliters of vinegar
So, in order to find the amount of soy sauce mixed with one ml of vinegar, we have to divide the amount of soy sauce by vinegar.
So,
Hence,
Bonnie is mixing 1.50 milliliter of soy sauce with 1 milliliter of vinegar
You might be interested in
Answer: She had 27.2% correct.
Step-by-step explanation: To find the percent, we can divide the part and whole then multiply the quotient by 100. After that we'll need to convert the percent to a decimal. In mathematical terms, the formula will be,
40% will be the part and 68% will be the whole.
<u><em>Step 1.</em></u>
Convert 40 into a decimal by dividing it by 100.
40 ÷ 100 = 0.4
<u><em>Step 2.</em></u>
Multiply 0.4 by 68.
0.4 × 68 = 27.2
Hence, she had 27.2% correct.
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that . Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that . In this equality we can perform a right multiplication by
and obtain
. Then, in the obtained equality we perform a left multiplication by P and get
. If we write
and
we have
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and . So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.