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yan [13]
2 years ago
14

You want to ship a box that is 10cm wide, 20 cm long and 5 cm high. What is the volume of the box?

Mathematics
2 answers:
asambeis [7]2 years ago
8 0

Answer:

......................... why do people do this?......................... I remember u u did this to me when i needed help so karma ........................................................................................................

Step-by-step explanation:

Molodets [167]2 years ago
7 0

Answer:

1000

Step-by-step explanation:

Volume = Length x Width x Height

20x10=200

200x5=1000

or you could do

10x5=50

20x50=1000

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kali has a 6 centimeter ladder and puts the base of the ladder 2 meters away from the tree how high into the tree will the lader
weqwewe [10]
I am going to assume ‘centimetre’ was a typo.

A^2 + B^2 = C^2

C^2 - A^2 = B^2

32 = B^2

Square root both sides

B = 5.7m high

The ladder will reach 5.7 meters high
5 0
3 years ago
If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=
Alexxx [7]
\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\

\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
3 years ago
40% of _ =$10:00<br><br> A $0.25<br> B $4.00<br> C $25.00<br> D $40.00<br> E none of these
True [87]
B $4.00
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4 0
3 years ago
Read 2 more answers
A polygon has its vertices at the following points.
seropon [69]

Answer:

isosceles trapezoid

Step-by-step explanation:

1) find the distance of the points

\sqrt{(x2-x1)^2 + (y2 -y1)^2}

AB = \sqrt{(4-2)^2 + (7-5)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}

BC = |7-4| = 3

CD = \sqrt{(9-7)^2 + (5-7)^2}= \sqrt{4+4} = \sqrt{8} = 2 \sqrt{2}

AD = |9-2| = 7

2) equation of the line that passes threw BC

y = 7

3) equation of the line that passes threw AD

y = 5

conclusion

the quadrilateral has two parallel sides and two congruent sides, so it is a isosceles trapezoid

3 0
2 years ago
The coordinates of the vertices of​ quadrilateral ABCD​ are A(−4, −1), B(−1, 2), C(5, 1), and D(1, −3).
serg [7]

Answer:

Sorry this is late, but it could help future people. Answers are in the picture.

Step-by-step explanation:

I took the quiz. :)

3 0
2 years ago
Read 2 more answers
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