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romanna [79]
3 years ago
5

Please help due tomorrow show work

Mathematics
1 answer:
Tju [1.3M]3 years ago
5 0
Little help a negative times a negative is always a positive so the answer is 24 and another hint is a negative times a positive is a negative
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A squirrel runs 23 feet down a hill to eat an acorn. Then it runs 23 feet up the hill. Write the integer that represents the squ
sdas [7]
The hill means it was already 23 feet above ground and he went and back up so the answer is 23 positive
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3 years ago
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Please help. Will give brainliest! No links or your account will be a banned
Firdavs [7]

Answer:

1. 18

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4. 84

Step-by-step explanation:

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2 years ago
The transitive property holds true for similar figures.<br><br> Always<br> Sometimes<br> Never
liubo4ka [24]

Answer: Always.


Step-by-step explanation:

The transitive property holds true for similar figures always because similar figures have similar shapes, the same angles and dimensions are proportional.

For example:- If figure 1 is similar to figure 2 then both have same shape and same angles and dimensions are proportional .

If figure 2 is similar to figure 3 then both have same shape and same angles and dimensions are proportional .

⇒ figure 1 is similar to figure 3 the both have same shape and same angles and dimensions are proportional as the figure 2 .

Thus the transitive property holds true for similar figures always.

5 0
3 years ago
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How can you break up the figure into familiar shapes to determine the area?
MissTica

Answer:

those are two different ways you can split the shape

8 0
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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
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
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