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11111nata11111 [884]
3 years ago
14

Solve by graphing u=v and 6u=2v-24

Mathematics
1 answer:
Vadim26 [7]3 years ago
8 0

Answer:

u=-6 \\ \\ v=-6

Step-by-step explanation:

In this exercise, we have two equations, namely:

u=v \ and \ 6u=2v-24

And we are asked to solve this problem by graphing. In this way, we can write a system of linear equations in two variables, but first of all, let's rewrite:

u=y \\ \\ v=x

Then:

\left\{ \begin{array}{c}y=x\\6y=2x-24\end{array}\right.

So here we have two lines.

The first one is:

\boxed{y=x}

This line passes through the origin and has a slope m=1

The second one is:

6y=2x-24 \\ \\ \therefore y=\frac{2x-24}{6} \\ \\ \therefore \boxed{y=\frac{1}{3}x-4}

This line has a slope m=\frac{1}{3} and cuts the y-axis at b=-4

By using graph tools, we get the graph shown below, then:

x=-6 \\ \\ y=-6 \\ \\ \\ Since \ u=y \ and \ v=x, then: \\ \\ u=-6 \\ \\ v=-6

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Answer:

Sally is not right

Step-by-step explanation:

Given the two sequences which have their respective n^{th} terms as following:

Sequence A. 3n - 2

Sequence B. 10 - 2n

As per Sally, there exists only one number which is in both the sequences.

To find:

Whether Sally is correct or not.

Solution:

For Sally to be correct, we need to put the n^{th} terms of the respective sequences as equal and let us verify that.

3n-2=10-2n\\\Rightarrow 3n+2n=10+2\\\Rightarrow 5n=12\\\Rightarrow n = \dfrac{12}{5}

When we talk about n^{th} terms, n here is a whole number not a fractional number.

But as per the statement as stated by Sally n is a fractional number, only then the two sequences can have a number which is in the both sequences.

Therefore, no number can be in both the sequences A and B.

Hence, Sally is not right.

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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