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Veseljchak [2.6K]
3 years ago
9

I was to lazy to type it out please let me know if you cant see it.

Mathematics
1 answer:
weqwewe [10]3 years ago
8 0

Not completely sure on this one but pretty sure it is 60 feet below sea level.
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Find the missing side of the triangle. Round your answer to the nearest tenth if necessary.
Soloha48 [4]

Answer:

Step-by-step explanation:

pythagoras theorem

a^2+b^2=c^2

8^2+x^2=10^2

64+x^2=100

x^2=100-64

x=\sqrt{36

x=6 km

5 0
3 years ago
Read 2 more answers
Which equation could you use to find 15 ÷ 3? 15 + 3 = 18 18 – 3 = 15 3 × 5 = 15 30 ÷ 2 = 15
kondor19780726 [428]
3 x 5 = 15 because the opposite of dividing is multiplying
4 0
3 years ago
Andrew has one book that is two and 378 inches thick and a second book that is 3.56 inches thick if he stacks the books about ho
Georgia [21]
2.378 + 3.56 = 5.938

two and 378 inches =  2.378 
8 0
3 years ago
Someone help me with my math homework pleaseee. Find the volumes of the pyramids and the height is 7cm for the first one.
gregori [183]
\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad 
\begin{cases}
B=\textit{area of the base}\\
h=height
\end{cases}

now, the first one, on the far-left.... can't see the height.. but I gather you do, now as far as its Base area, well, the bottom is just a 12x12 square, so the area of its base is just 12*12


now, the middle pyramid, has a height of 6, the base is also a square, 8x8, so the Base area is just 8*8

now the last one on the far-right

has a height of 8, the Base is a Hexagon, with sides of 6

\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{4}ns^2cot\left( \frac{180}{n} \right)\qquad 
\begin{cases}
n=\textit{number of sides}\\
s=\textit{length of one side}\\
\frac{180}{n}=\textit{angle in degrees}\\
----------\\
n=6\\
s=6
\end{cases}\\\\\\ A=\cfrac{1}{4}\cdot 6\cdot 6^2\cdot cot\left( \frac{180}{6} \right)
5 0
3 years ago
Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S
Sophie [7]

Answer:

Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that A=J^{-1}AJ. Thus, A↔A.

<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that A=P^{-1}BP. In this equality we can perform a right multiplication by P^{-1} and obtain AP^{-1} =P^{-1}B. Then, in the obtained equality we perform a left multiplication by P and get PAP^{-1} =B. If we write Q=P^{-1} and Q^{-1} = P we have B = Q^{-1}AQ. Thus, B↔A.

<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have A=P^{-1}BP and from B↔C we have B=Q^{-1}CQ. Now, if we substitute the last equality into the first one we get

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
3 years ago
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