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3 years ago

What is the word that means the result of a transformation ?

Mathematics

1 answer:

IrinaVladis [17]3 years ago

8 0

Answer: The result of a transformation in math is called "image" not "Pre-image."

Hope this helps..... Stay safe and have a Merry Christmas!!!!!!! :D

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Using substitution x+3y=3 3y-2x=12

frez [133]

Step-by-step explanation:

Given :-

x + 3y = 3
3y - 2x = 12

And we need to solve the equation using Substituting method . So on taking the first equation ,

$\rm\implies x + 3y = 3 \\\\\rm\implies x = 3 - 3y$

Put this value in (ii) :-

$\rm\implies 3y - 2x = 12 \\\\\rm\implies 3y - 2(3-3y)=12\\\\\rm\implies 3y -6-6y = 12 \\\\\rm\implies -3y = 18 \\\\\rm\implies y = -6$

Put this Value in (I) :-

$\\\\\rm\implies x = 3-3*-6 \\\\\rm\implies x = 3+18 \\\\\rm\implies x = 21$

Hence the Value of x is 21 and y is (-6) .

4 0

3 years ago

Read 2 more answers

Please help me with this please and thank you please actually help me

fgiga [73]

Answer:

-5, -8, -11

Step-by-step explanation:

it every -3

5 0

2 years ago

Read 2 more answers

A value meal package at Ron's Subs consists of a drink, a sandwich, and a bag of chips. There are 5 types of drinks to choose fr

Lynna [10]

Answer: 90 different meal packages.

Step-by-step explanation:

Given data:

Types of drinks available = 5

Types of sandwiches available = 6

Types of chips available = 3.

Solution:

No of meal packages possible for this combination

= 6*5*3

= 90

A total of 90 different meal packages would be gotten:

8 0

2 years ago

I need help... on my homework it says to write each expression in radical form, or write each radical and exponential form . For

pantera1 [17]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt{13}\implies \sqrt[2]{13^1}\implies 13^{\frac{1}{2}}$

5 0

3 years ago

If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b

MAXImum [283]

We have

$\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}$

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0$

The rational term vanishes as x gets arbitrarily large, so we can ignore that term, leaving us with

$\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0$

and this happens if a = 1 and b = -1.

To confirm, we have

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0$

as required.

3 0

3 years ago