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a_sh-v [17]
2 years ago
11

The regular price of a digital camera was 250$. Tristan bought the digital camera at a discount of 40%. How much did Tristan pay

for the digital camera?
Mathematics
2 answers:
Bingel [31]2 years ago
7 0

Answer:

Step-by-step explanation:

Sale price = 40% of $250

sale price = 40/100 x 250

sale price = 0.4 x 250

sale price = $100

New price = original price - sale price

new price = 250 - 100

new price = $150

enot [183]2 years ago
3 0

Answer:

$150

Step-by-step explanation:

convert the percent to decimal, 0.40, then multiply it by 250 to get 100, then subtract, 250-100=$150

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Answer:

Step-by-step explanation:

From the picture attached,

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1). Arrow starting with a solid point from x = -2                   x ≤ -2

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7 0
3 years ago
Subtract using the number line -3/5 - (-2/5) (picture provided)
lara31 [8.8K]

Answer:

<h2>-1/5</h2>

Step-by-step explanation:

Starting at -3/5 you need to subtract -2/5.

When you subtract a negative number, the sign flips and you add instead. Often, you are taught that two similar signs means add and two opposite signs mean minus.

<em>+ + = +</em>

<em>- - = -</em>

<em>+ - = -</em>

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<em />

This means -3/5 - -2/5 becomes -3/5 + 2/5.

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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
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\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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