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dalvyx [7]
3 years ago
13

Sequence of transformation that take the graph y=x^2 to y=-2(x-5)^2+4

Mathematics
1 answer:
Lisa [10]3 years ago
3 0

Answer:

(x-5) so translated 5 units to the right

Multiplied with 2, p vertically compressed

+4 means translated 4 units up

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Catron evaluates the expression (negative 9) (2 and two-fifths) using the steps below.
Vlad [161]

Answer:

Catron's error is

"She did not follow order of operations"

Step-by-step explanation:

Catron evaluates the expression (negative 9) (2 and two-fifths)

That expression can be written as below

(-9)(2\frac{2}{5})

Catron's error is

"She did not follow order of operations"

The corrected steps are

Step1: Given expression is (-9)(2\frac{2}{5})

Step2: Convert mixed fraction into improper fraction

(-9)(2\frac{2}{5})=(-9)(\frac{12}{5})

Step3: Multiplying the terms

(-9)(2\frac{2}{5})=-7\frac{-108}{5}

Therefore solution (-9)(2\frac{2}{5})=-7\frac{-108}{5}

6 0
3 years ago
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What is the range of a sine function?
vesna_86 [32]

Answer:

All real numbers between -1 and 1, including -1 and 1

Step-by-step explanation:

7 0
3 years ago
Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S
Sophie [7]

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 A=J^{-1}AJ. 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 A=P^{-1}BP. In this equality we can perform a right multiplication by P^{-1} and obtain AP^{-1} =P^{-1}B. Then, in the obtained equality we perform a left multiplication by P and get PAP^{-1} =B. If we write Q=P^{-1} and Q^{-1} = P we have B = Q^{-1}AQ. 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 A=P^{-1}BP and from B↔C we have B=Q^{-1}CQ. Now, if we substitute the last equality into the first one we get

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
3 years ago
What is equivalent to 5-squared3
pentagon [3]

Answer: 125

Step-by-step explanation:

5^3=5*5*5=125

-5^3=(-5)*(-5)*(-5)=25*(-5)=-125

5 0
3 years ago
The range of F(x) = logb x is the set of all real numbers. True or false
LiRa [457]

The range of F(x) = logb x is True

5 0
3 years ago
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