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2 years ago

Which answer choice is equivalent to -65-12?

Mathematics

2 answers:

elena-14-01-66 [18.8K]2 years ago

6 0

Answer:

Taht problem equals -77 so I would say d, good luck!

Alex17521 [72]2 years ago

5 0

I would say the answer would be d !

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2a^2-5a=3 solve by factoring

Amiraneli [1.4K]

Answer:

a=−1/2, 3

3Step-by-step explanation:

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3 years ago

A candy company has orders for chocolate bars feom 5 different stores. Each order dontains 45 chocolate bars. Use he distrubutiv

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2 years ago

Read 2 more answers

What quadrant is (20, 10) in??

dalvyx [7]

Answer:

(20,10) is gonna be in The first quadrant

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2 years ago

The plans for a rectangular deck call for the width to be 10 feet less than the length. Sam wants the deck to have an overall pe

sergeinik [125]

Length of deck is 40 feet

<h3><u><em>Solution:</em></u></h3>

Sam wants the deck to have an overall perimeter of 60 feet

Perimeter of rectangular deck = 60 feet

Let "L" be the length of rectangle and "W" be the width of rectangle

Given that plans for a rectangular deck call for the width to be 10 feet less than the length

Width = length - 10

W = L - 10 ------ eqn 1

<em><u>The perimeter of rectangle is given as:</u></em>

perimeter of rectangle = 2(length + width)

Substituting the known values we get,

60 = 2(L + L - 10)

60 = 2(2L - 10)

60 = 4L - 20

80 = 4L

L = 20

Thus the length of deck is 20 feet

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3 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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3 years ago