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3 years ago

Agree or desagree? short question'

Mathematics

2 answers:

Mekhanik [1.2K]3 years ago

7 0

Answer:

Disagree

Step-by-step explanation:

tatiyna3 years ago

5 0

Disagree because it wouldn’t make sense

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4 years ago

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Which expression is equivalent to

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ITS D ITS NOT THERE

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3 years ago

Help Please.. Can't figure it out

Alex787 [66]

Answer:

The answer is option 3.

Step-by-step explanation:

The area of triangle formula is A = 1/2×base×height. Then substitute the following values into the formula :

$area = \frac{1}{2} \times base \times height$

Let base = 10,

Let height = 12,

$area = \frac{1}{2} \times 10 \times 12$

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4 years ago

3 in.

gtnhenbr [62]

I would help but i don’t see a figure

7 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 $x=\pm 2$

5 0

4 years ago

Read 2 more answers

Agree or desagree? short question'​

Agree or desagree? short question'