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satela [25.4K]
3 years ago
7

Quest-Room ist die beste Unternehmung für Gruppen in Köln und liegt nur wenige Minuten vom Hauptbahnhof und dem Kölner Dom entfe

rnt. Ob ihr aus dem Gefängnis fliehen oder der sinkenden Titanic entkommen wollt liegt in eurer Hand, weil bei uns habt ihr die Möglichkeit zwischen vielen verschiedenen Escape-Rooms (Puzzle und Rätselräumen) zu wählen. Besucht uns entweder auf unserer Website oder vor Ort Habsburgerring 3.

Mathematics
1 answer:
kramer3 years ago
8 0
Idk...........................................................................................
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Translate to an equation and solve: The sum of two-fifths and f is one-half.<br> Provide<br> f=
SCORPION-xisa [38]

Answer:

9/10

1/2+2/5

Since these fractions have different denominators, we need to find the least common multiple of the denominators.The least common multiple of 2 and 5 is 10, so we need to multiply to make each of the denominators = 10

1/2 ∗ 5/5 = 5/10

5/2∗ 2/2= 4/10

Since these fractions have the same denominator, we can just add the numerators

5/10+ 4/10 = 9/10

3 0
3 years ago
If a number is divisible by both 2 and 3 then we can say the number is divisible by A. 5. B. 2. C. 4. D. 6.
Elza [17]
I believe it is d 6. you can divide 6 by both 3 and 2.
7 0
3 years ago
Read 2 more answers
If a cylender has a volume of 340ft and a radius of three ft what is the height
rewona [7]

Answer:

h = 12.03

Substitute 22÷7 for pi.

h = V ÷ πr²

h = 340 ÷ (22÷7) × 3²

   ≈ 12.03

7 0
2 years ago
498,147,846,267<br>+ 32,870,714,538<br>​
san4es73 [151]

Answer:

531018560805

Step-by-step explanation:

4 0
3 years ago
I really am confused on how to write a proof.
kvasek [131]

Since ABCD is a parallelogram, the opposite sides will be parallel and equal,

\begin{gathered} AB=CD \\ BC=AD \end{gathered}

Consider that AC acts as a transversal to the parallel lines AB and CD, so we can write,

\begin{gathered} \angle CAD=\angle ACB\text{ (Alternate Interior Angles)} \\ BC=AD\text{ (Opposite sides of parallelogram)} \\ \angle ADB=\angle CBD\text{ (Alternate Interior Angles)} \end{gathered}

So by the ASA criteria, the triangle AED is congruent to the triangle CEB,

Then the corresponding parts of the triangles will be equal,

\begin{gathered} AE=CE \\ BE=DE \end{gathered}

Hence Proved.

8 0
1 year ago
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