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3 years ago

True or false: 29 < 5

Mathematics

1 answer:

elena55 [62]3 years ago

5 0

Answer:

False

Step-by-step explanation:

29 is more than 5

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

If a = 2, b = 3 and c = 4, what is the numerical value of the expression (b-c)^2 + a(b+c)?

wel

Answer:

Step-by-step explanation:

(3 - 4)^2 + 2(3 + 4)

(-1)^2 + 2(7)

1 + 14

3 0

3 years ago

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EFGH is a rhombus. Given EG = 16 and FH = 12, what is the length of one side of the rhombus?

Black_prince [1.1K]

Answer:

Step-by-step explanation:

It is given that EFGH is a rhombus and EG = 16 and FH = 12.

We know that the diagonals of the rhombus are the perpendicular bisectors, therefore OF=6, OH=6, OE=8 and OG=8.

Now, using the Pythagoras theorem in ΔOFG, we have

$(FG)^{2}=(OG)^{2}+(OF)^{2}$

$(FG)^{2}=(8)^{2}+(6)^{2}$

$(FG)^2=64+36$

$FG=\sqrt{100}$

$FG=10$

Thus, the value of the side of the rhombus will be 10 units.

Hence, option C is correct.

3 0

4 years ago

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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 $x=\pm 2$

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4 years ago

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