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Mariulka [41]
1 year ago
6

What is the domain, range, and function? {(-3, 3), (1, 1), (0, -2), (1,4), (5, -1)}

Mathematics
1 answer:
wolverine [178]1 year ago
8 0

The domain are all the inputs, that is; the x-values

Domain = { -3, 0, 1, 5}

The range are all the output, that is the y-values

Range = { 3, 1, -2, 4, -1}

This is just a relation and not a function, as we have more than 2 same value of x

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What transformations were applied to ABCD to obtain A'B'C'D?
ch4aika [34]

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Step-by-step explanation:

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Is 1.6 more than, less than, or equal to one and a half? Explain why
hram777 [196]
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4 0
3 years ago
Read 2 more answers
The population of Rwanda in 2016 was about 11.8 million. The population tends to increase by about 2% per year. Which equation d
natita [175]

Answer:

B - P=11.8(1.02)^t

Step-by-step explanation:

An exponential function is defined as:

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5 0
3 years ago
(1 point) Find two unit vectors orthogonal to a=⟨4,5,0⟩a=⟨4,5,0⟩ and b=⟨0,1,−3⟩b=⟨0,1,−3⟩ Enter your answer so that the first no
Anni [7]

Answer:

 First vector

V=+ \dfrac{-15i+12j+4k}{\sqrt{385}}

Second vector

V=- \dfrac{-15i+12j+4k}{\sqrt{385}}

Step-by-step explanation:

Given that

Vector a ,  a= 4 i + 5 j + 0 k

Vector b , b= 0 i + 1 j  - 3 k

The orthogonal vector V is given as

V = a x b

V is the cross product of the above two vector which is orthogonal to the vectors a and b.

V=\begin{vmatrix}i & j & k\\ 4 & 5 & 0\\  0& 1 &-3 \end{vmatrix}\\V=i(-15-0)-j(-12-0)+k(4-0)\\V=-15i+12j+4k

The unit vectors

V=\pm \dfrac{-15i+12j+4k}{\sqrt{15^2+12^2+4^2}}\\\\V=+\dfrac{-15i+12j+4k}{\sqrt{385}}\\V=-\dfrac{-15i+12j+4k}{\sqrt{385}}

First vector

V=+ \dfrac{-15i+12j+4k}{\sqrt{385}}

Second vector

V=- \dfrac{-15i+12j+4k}{\sqrt{385}}

5 0
3 years ago
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