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3 years ago

Y= 4x+1:

Mathematics

2 answers:

FrozenT [24]3 years ago

5 0

Answer:

Y=IS NOT A FUNCTION Y IN TERMS OF ONE VARIABLE

THERE IS NO SOLUTION

Step-by-step explanation:

Troyanec [42]3 years ago

3 0

You have to put the value of x in the equation to get the answer
Y = 4x+1
Y = 4(2)+1
Y= 8+1
Y= 9
(2,9)

Y= 4x+1
Y= 4(6)+1
Y= 24+1
Y= 25
(6,25)

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Which expression is equivalent to 200? O2./10 1012 10./20 1002

solong [7]

Hey there!

We know options. B and D aren't going to be your answer because they are way too big to be equivalent to 200

In order for you to find out which one is equivalent to 200 you have to calculate you late the options until you get to 200 and then you could do process of elimination!

It can't be A because 2/10 = 0.20 aka 20%

It can't be option. C because 10/20 = 0.50 aka 50%

So none of the above would be your answer

The correct answer would most likely be: 10√2

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

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3 years ago

The perimeter of a triangle is 64'. If one side is x feet, the second side is 3x + 1 feet and the third

Dominik [7]

One side is 9.
second side (3x+1) is 28.
third side (4x-9) is 27.

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3 years ago

What is the constant in the following equation? 12 = 3k

stiks02 [169]

Answer:

k = 4

Step-by-step explanation:

$12=3k\\\frac{12}{3}=\frac{3k}{3} \\4=k$

5 0

3 years ago

An=an-1+6an-2 n>2 a0=3 a1=6

Svetach [21]

$\begin{cases}a_0=3\\a_1=6\\a_n=a_{n-1}+6a_{n-2}&\text{for }n>2\end{cases}$

Let the generating function for $a_n$ be

$A(x)=\displaystyle\sum_{n\ge0}a_nx^n$

Multiplying both sides by $x^{n-2}$

$a_nx^{n-2}=a_{n-1}x^{n-2}+6a_{n-2}x^{n-2}$

Summing both sides over the non-negative integers greater than or equal to 2 gives

$\displaystyle\sum_{n\ge2}a_nx^{n-2}=\sum_{n\ge2}a_{n-1}x^{n-2}+6\sum_{n\ge2}a_{n-2}x^{n-2}$
$\displaystyle\frac1{x^2}\sum_{n\ge2}a_nx^n=\frac1x\sum_{n\ge1}a_nx^n+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(\sum_{n\ge0}a_nx^n-a_0-a_1x\right)=\frac1x\left(\sum_{n\ge0}a_nx^n-a_0\right)+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)$
$A(x)=\dfrac{3x+3}{1-x-6x^2}$
$A(x)=\dfrac3{5(1+2x)}+\dfrac{12}{5(1-3x)}$

For $|x|$ , the two series converge to

$\displaystyle A(x)=\frac35\sum_{n\ge0}(-2x)^n+\frac{12}5\sum_{n\ge0}(3x)^n$
$\implies A(x)=\displaystyle\sum_{n\ge0}\dfrac{3(-2)^n+12(3)^n}5x^n$
$a_n=\dfrac{3(-2)^n+12(3)^n}5=\dfrac{3(-2)^n+4(3)^{n+1}}5$

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3 years ago

Henry is 3 times older than Julia. Rhianna is 1/2 as old as Julia. What formula shows the relationship between Henry and Rhianna

Bess [88]

H=3J, R=(1/2)J; times the R equation by 6 to get 6R=3j then substitute into the H equation to get H=6R

3 0

3 years ago