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3 years ago

HELLPPP I need help asappp

Mathematics

1 answer:

sattari [20]3 years ago

3 0

Answer:

(8,5)

Step-by-step explanation:

x=8

5*8-2y=30

40-2y=30 . subtract 40 from both sides

-2y=-10 . divide both sides by-2

y=5

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Please Help!!

Genrish500 [490]

Given

$a\sqrt{x+b}+c=d$

we have

$\sqrt{x+b}=\dfrac{d-c}{a}$

Squaring both sides, we have

$x+b=\dfrac{(d-c)^2}{a^2}$

And finally

$x=\dfrac{(d-c)^2}{a^2}-b$

Note that, when we square both sides, we have to assume that

$\dfrac{d-c}{a}>0$

because we're assuming that this fraction equals a square root, which is positive.

So, if that fraction is positive you'll actually have roots: choose

$a=1,\ b=0,\ c=2,\ d=6$

and you'll have

$\sqrt{x}+2=6 \iff \sqrt{x}=4 \iff x=16$

Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose

$a=1,\ b=0,\ c=10,\ d=4$

and you'll have

$\sqrt{x}+10=4 \iff \sqrt{x}=-6$

Squaring both sides (and here's the mistake!!) you'd have

$x=36$

which is not a solution for the equation, if we plug it in we have

$\sqrt{x}+10=4 \implies \sqrt{36}+10=4 \implies 6+10=4$

Which is clearly false

7 0

3 years ago

Your friend claims that when a polynomial function has a leading coeffcient of 1 and the coefficients are all integers , every p

sammy [17]

Using the rational root theorem, it is found that your friend is correct.

<h3>What is the rational root theorem?</h3>

It is a theorem that states that for a polynomial with integer coefficients, with q being the factors of the leading coefficient and p being the factors of the constant, every <u>possible rational root</u> is the format $\frac{p}{q}$ .

In this problem:

The leading coefficient is 1, hence it's only factor is $q = 1$ , thus guaranteeing that every possible rational zero is an integer, which means that your friend is correct.

To learn more about the rational root theorem, you can take a look at brainly.com/question/10937559

8 0

2 years ago

For what interval is the function f(x) = (20 + \sqrt{x}) / (\sqrt{20 + x}) continuous?

konstantin123 [22]

Try this solution:
1. according to the condition
$\left \{ {{x \geq 0} \atop {20+x\ \textgreater \ 0}} \right. \ =\ \textgreater \ \ x \geq 0.$
2. for more details see the attached graph.

Answer: [0;+oo)