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2 years ago

Question 16 of 17

Mathematics

1 answer:

FrozenT [24]2 years ago

3 0

The answer is B. It is a relation and also a function at the same time because it satisfies a specific domain

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Need help plz help

torisob [31]

Answer:

A. Square for Ape.x

Step-by-step explanation:

8 0

3 years ago

A is a figure with ten sides and ten interior angles.

wel

Answer:

not enough information to form an answer

Step-by-step explanation:

and what is ur profile pic lol

7 1

3 years ago

Read 2 more answers

Simplify using the laws of exponents (4^3)^-2 × (2^3)^4 ×(8/15)^-2

katrin2010 [14]

Answer:

$\dfrac{225}{64}$

Step-by-step explanation:

$(4^3)^{-2} \times (2^3)^4 \times (\dfrac{8}{15})^{-2} =$

$= (2^2)^{-6} \times 2^{12} \times (\dfrac{15}{8})^{2}$

$= 2^{-12} \times 2^{12} \times \dfrac{225}{64}$

$= \dfrac{225}{64}$

8 0

3 years ago

Negative eleven times positive 8 equals

Alex73 [517]

Answer:

-11 x 8 = -88

Step-by-step explanation:

negative times positive = negative

negative times negative = positive

positive times positive = positive

7 0

3 years ago

1. Find a vector in R^3 which is not in the span of the set S-((1,2, -2), (2, -1, 1). Explain.

pav-90 [236]

Answer:

Step-by-step explanation:

A vector in R^3 which is not in the span of the set S {(1,2,-2) and (2,-1,1)

If a vector is in the span it can be represented as a linear combination of these two vectors

Let S1 = (1,2,-2) and S2 = (2,-1,1)

i.e. any vector which is of the form

$\alpha S1+\beta S2\\=(\alpha +2\beta,2\alpha -\beta,-2\alpha +\beta)$

Where alpha and beta are any real numbers

Any vector not in this form will not be in the span

i.e. say if alpha = beta =1,

then spanned vector = (3,1,-1)

If we change one coordinate alone say

(3,0,-1) this cannot be represented as a linear combination hence this would be the answer.

8 0

3 years ago