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3 years ago

Helpppppppppppppppppppppppppp me owo plzzzzzzzzzz

Mathematics

2 answers:

Tanya [424]3 years ago

8 0

Answer:

Just concentrate you can do it

Step-by-step explanation:

Inga [223]3 years ago

6 0

Answer:

$(-8x^2+7x-7)-(10x^2-10x+2)\\-8x^2+7x-7-10x^2+10x-2\\-8x^2-10x^2+7x+10x-7-2\\-18x^2+17x-9$

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The output is the cube of the input

MakcuM [25]

The output is the cute of the input is written as y = x³.

5 0

3 years ago

Use properties of limits and algebraic methods to find the limit, if it exists. (If the limit is infinite, enter '[infinity]' or

soldi70 [24.7K]

Answer:

The limit of this function does not exist.

Step-by-step explanation:

$\lim_{x \to 3} f(x)$

$f(x)=\left \{ {{9-3x} \quad if \>{x \>< \>3} \atop {x^{2}-x }\quad if \>{x\ \geq \>3 }} \right.$

To find the limit of this function you always need to evaluate the one-sided limits. In mathematical language the limit exists if

$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) =L$

and the limit does not exist if

$\lim_{x \to a^{-}} f(x) \neq \lim_{x \to a^{+}} f(x)$

Evaluate the one-sided limits.

The left-hand limit

$\lim_{x \to 3^{-} } 9-3x= \lim_{x \to 3^{-} } 9-3*3=0$

The right-hand limit

$\lim_{x \to 3^{+} } x^{2} -x= \lim_{x \to 3^{+} } 3^{2}-3 =6$

Because the limits are not the same the limit does not exist.

8 0

3 years ago

I don't get these what Do I do

max2010maxim [7]

Answer:

No idea

Step-by-step explanation:

3 0

2 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

6 0

3 years ago

Is 9.707 greater than 9.717

stepan [7]

No 9.717 is greater than 9.707 because 9.717 would be seven hundred seven thousandths and 9.717 would be seven hundred seven teen hundredths.

8 0

3 years ago