If five times the successor of a number is added to an original number the sum is 83

Mathematics

1 answer:

Alexus [3.1K]3 years ago

8 0

Answer:

Let the original number be x

Successor is defined as the number which comes immediately after a particular number.

also, the successor of a whole number is the number obtained by adding 1 to that number.

Then, the successor of a number x is, x+1

As per the given condition :

we have;

$5(x+1)+x =83$

Using distributive property on LHS (i.e, $a\cdot(b+c)=a\cdot b+a\cdot c$ )

Then, we have

5x+5+x=83

Combine like terms;

6x+5=83

Subtract 5 from both the sides we get;

6x+5-5=83-5

Simplify:

6x=78

Divide both side by 6,

$\frac{6x}{6} =\frac{78}{6}$

Simplify:

x =13

Therefore, the original number x is, 13

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I don’t know how to do this

stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as $x$ approaches the edges. For example,

$g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}$

has two pieces, $2x+5$ and $x^2-10$ , both of which are continuous by themselves on the provided intervals. In order for $g$ to be continuous everywhere, we need to have

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)$

By definition of $g$ , we have $g(-3)=2(-3)+5=-1$ , and the limits are

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1$

$\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1$

The limits match, so $g$ is continuous.

For the others: Each of the individual pieces of $f,h$ are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.