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Alecsey [184]
3 years ago
8

Anybody know How too do this stuff

Mathematics
1 answer:
Stells [14]3 years ago
8 0
Find the greatest common factor of both the numbers
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Cavalieri's principle states that two solids with equal heights and base areas have equal volumes. A.True B.False
Butoxors [25]

Answer:

So, the following statement is a TRUE statement.

Step-by-step explanation:

In geometry Cavalieri's priciple states that:

Cavalieri's Principle: If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

Hence, the given statement:

Cavalieri's principle states that two solids with equal heights and base areas have equal volumes is a TRUE statement.

8 0
3 years ago
Read 2 more answers
2x+6=-72 <br> what does x equal
shepuryov [24]

Answer:

- 39

Step-by-step explanation:

Step 1:

2x + 6 = - 72         Equation

Step 2:

2x = - 78      Subtract 6 on both sides

Step 3:

x = - 78 ÷ 2         Divide

Answer:

x = - 39

Hope This Helps :)

8 0
4 years ago
Read 2 more answers
HELP AGAIN STILL BEING TIMED
Rama09 [41]

Answer:

C

Step-by-step explanation:

6 0
3 years ago
What is another way to write this number 300+70+5/10+8/100
Kobotan [32]
Hello!

Your answer is 370 \frac{29}{50}
7 0
3 years ago
Read 2 more answers
Sin 0 = 0.731353701, 0º PLEASE HELP!!
Snezhnost [94]

Answer:

The two solutions for \sin \theta = 0.731353701 are \theta_{1} \approx 47^{\circ} and \theta_{2}\approx 133^{\circ}, respectively.

Step-by-step explanation:

The sine function is positive in the first and second quadrants, that is, 0^{\circ}< \theta < 180^{\circ}. Then, there are two solutions for the given value by means of the following inverse trigonometrical function:

\theta = \sin^{-1} 0.731353701

\theta_{1} \approx 47^{\circ} and \theta_{2}\approx 133^{\circ}

The two solutions for \sin \theta = 0.731353701 are \theta_{1} \approx 47^{\circ} and \theta_{2}\approx 133^{\circ}, respectively.

7 0
3 years ago
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