Which equation has no solution?

Mathematics

1 answer:

steposvetlana [31]3 years ago

3 0

Answer:

B i am pretty sure

Step-by-step explanation:

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9) 15k + 20 – 10k = 10

Mashutka [201]

Answer:

k=-2

Step-by-step explanation:

You have to bring k itself and make it to a coefficient of 1 using algebra.

6 0

3 years ago

Find the common difference of the sequence shown. 1/2, 1/4, 0, ... -1/8 -1/4 -1/2

DENIUS [597]

Common difference in Arithmetic sequences is defined as the fixed amount referring to the fact that the difference between two successive terms yields the constant value that was added. In order to know the common difference of the given sequence above, we just have to deduct 1/2 from 1/4 and the answer is -1/4. Therefore, the correct answer would be the third option. Hope this answer helps.

8 0

3 years ago

State the vertical asymptote of the rational function. f(x) =((x-9)(x+7))/(x^2-4)

Lubov Fominskaja [6]

$D:x^2-4\not=0\\
D:x^2\not=4\\
D:x\not=-2 \wedge x\not =2\\\\
\displaystyle
\lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\
\dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty
$

$\displaystyle
\lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\
\dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty\\$

So, the vertical asymptotes are $x=\pm 2$