Don’t put no links just put a answer now can someone help me

Mathematics

1 answer:

Harlamova29_29 [7]2 years ago

4 0

Nooooooooooooooooooooooooo

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Can someone help me please

Fudgin [204]

Answer:

The answer is D

Step-by-step explanation:

1/8 is the equivalent to 0.12

square root of 0.02 is 0.14

18% is equal to 0.18

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2 years ago

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How do you do this, can someone help please?

igor_vitrenko [27]

Answer:

your answer is C.

Step-by-step explanation:

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3 years ago

Farrah borrowed $155 from her brother. She has already paid back $15. She plans to pay back $35 each month until the debt is pai

Sergeeva-Olga [200]

Answer:

4 months

Step-by-step explanation:

if she paid the 15 already subtract it from the 155. Once you that you get 140 and if you break that inton groups of 35 *division* than you get 4. As in 4 months

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Please HELP!!<br> Which linear inequality is represented by the graph

garri49 [273]

Answer:

The answer should be y ≤ 1/3x - 1 I believe

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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]

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

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

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2 years ago