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trapecia [35]
3 years ago
13

The unit is geometry

Mathematics
1 answer:
Tcecarenko [31]3 years ago
5 0
Question 1. 90.01

Question 2. No They Dont Have Enough Clay Because Your Taken 3inches Away
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Evaluate the expression 3.14(a2 + ab) when a = 4 and b = 3. (Input decimals only, such as 12.71, as the answer.
kondaur [170]

Answer:

87.92

Step-by-step explanation:

3.14(a²+ab)=

<em>Plugging in values for a and b</em>

3.14(4²+4×3)=

3.14(16+12)=

3.14(28)=

87.92

6 0
3 years ago
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Please help!! Giving brainliest and extra points! (Click on photo)
tamaranim1 [39]

Answer:

B, C and I think D

Step-by-step explanation:

5 0
3 years ago
Expand the expression using the distributive property. 2(5x – 3)​
riadik2000 [5.3K]

Answer:

10x-6

Step-by-step explanation:

4 0
3 years ago
Average winter temperatures in 45 states were recorded and divided by geographic region. Choose True or False for each statement
san4es73 [151]

Answer: SU, TU, ST is the right answer

Step-by-step explanation: In a triangle

The largest side and the largest angle are opposite to each other

The shortest side and the shortest angle are opposite to each other.

So using the above axioms we can observe the triangle.

The largest angle is m∠U = 80° so the side opposite to it is the largest i.e. ST is the largest

The smallest angle is m∠T = 25° so the shortest side is SU

The third side will be the middle side as the angle ∠S is greater than 25° and less than 80°

So the sides in order from least to greatest are: SU, TU, ST

5 0
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}

6 0
3 years ago
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