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kodGreya [7K]
3 years ago
12

PleAse help and show how you got it if you don’t mind

Mathematics
1 answer:
Lorico [155]3 years ago
3 0
1)

\bf \textit{surface area of a cube}\\\\
s=6x^2\quad 
\begin{cases}
x=\textit{length of a side}\\
---------\\
s=150
\end{cases}\implies 150=6x^2\implies \cfrac{150}{6}=x^2
\\\\\\
25=x^2\implies \sqrt{25}=x\implies \boxed{5=x}

3)

\bf \textit{volume of a square pyramid}\\\\
V=\cfrac{b^2h}{3}\quad 
\begin{cases}
b=\textit{base's side}\\
h=height\\
-------\\
h=3\\
V=25
\end{cases}\implies 25=\cfrac{b^2\cdot 3}{3}\implies 25=b^2\cdot \cfrac{3}{3}
\\\\\\
25=b^2\cdot 1\implies 25=b^2\implies \sqrt{25}=b\implies \boxed{5=b}
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Answer:

A.

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Step-by-step explanation:

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Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

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Let us try to find out \angle E and \angle DFE. After that we will compare them.

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<u>Finding </u>\angle E<u>:</u>

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i.e. external \angle DFG = \angle E + \angle EDF

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Comparing equations (1) and (2):

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\angle DFE = \angle E =90-\dfrac{y}{2}

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