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3 years ago

Find the solution(s) of the following equation. x2 = 36/49

Mathematics

1 answer:

abruzzese [7]3 years ago

6 0

Answer:

$x=-\frac{6}{7} , x=\frac{6}{7}$

Step-by-step explanation:

We begin with the equation $x^2=\frac{36}{49}$

In order to get x on the left side, we need to square root both sides. As we are taking the square root of each sides, one of the sides must have a ± added to it in order to make sure that both of the solutions are found.

$\sqrt{x^2} =$ ± $\sqrt{\frac{36}{49} }$

We can then simplify this

$x=$ ± $\frac{\sqrt{36} }{\sqrt{49} }$

And once more

$x=$ ± $\frac{6}{7}$

Which means that our solutions are

$x=-\frac{6}{7} , x=\frac{6}{7}$

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Can someone please help me?

kvasek [131]

now, let's recall the rational root test, check your textbook on it.

so p = 18 and q = 1

so all possible roots will come from the factors of ±p/q

now, to make it a bit short, the factors are loosely, ±3, ±2, ±9, ±1, ±6.

recall that, a root will give us a remainder of 0.

let us use +3.

$\bf x^4-7x^3+13x^2+3x-18 \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{r|rrrrr} 3&1&-7&13&3&18\\ &&3&-12&3&18\\ \cline{1-6} &1&-4&1&6&0 \end{array}\qquad \implies (x-3)(x^3-4x^2+x+6)$

well, that one worked... now, using the rational root test, our p = 6, q = 1.

so the factors from ±p/q are ±3, ±2, ±1

let's use 3 again

$\bf \begin{array}{r|rrrrr} 3&1&4&1&6\\ &&3&-3&-6\\ \cline{1-5} &1&-1&-2&0 \end{array}\qquad \implies (x-3)(x-3)(x^2-x-2)$

and of course, we can factor x²-x-2 to (x-2)(x+1).

(x-3)(x-3)(x-2)(x+1).

7 0

3 years ago

Y = (x)^(1/9)*Find f(x) when x=(1/2)Round your answer to the nearest thousandth.

Verizon [17]

We are given a function f ( x ) defined as follows:

$y=f(x)=x^{\frac{1}{9}}$

We are to determine the value of f ( x ) when,

$\times\text{ = }\frac{1}{2}$

In such cases, we plug in/substitue the given value of x into the expressed function f ( x ) as follows:

$y\text{ = f ( }\frac{1}{2}\text{ ) = (}\frac{1}{2})^{\frac{1}{9}}$

We will apply the power on both numerator and denominator as follows:

$f(\frac{1}{2})=\frac{1^{\frac{1}{9}}}{2^{\frac{1}{9}}}\text{ = }\frac{1}{2^{\frac{1}{9}}}$

Now we evaluate ( 2 ) raised to the power of ( 1 / 9 ).

$f\text{ ( }\frac{1}{2}\text{ ) = }\frac{1}{1.08005}$

Next apply the division operation as follows:

$f\text{ ( }\frac{1}{2})\text{ = }0.92587$

Once, we have evaluated the answer in decimal form ( 5 decimal places ). We will round off the answer to nearest thousandths.

Rounding off to nearest thousandth means we consider the thousandth decimal place ( 3rd ). Then we have the choice of either truncating the decimal places ( 4th and onwards ). The truncation only occurs when (4th decimal place) is < 5.

However, since the (4th decimal place) = 8 > 5. Then we add ( 1 ) to the 3rd decimal place and truncate the rest of the decimal places i.e ( 4th and onwards ).

The answer to f ( 1 / 2 ) to the nearest thousandth would be: