So if we think of a test tube, it looks sort of like a cylinder. This means that its cross-section would be a circle. To find out how many turns a piece of thread would make around the test tube, we need to find the circumference of the test tube, then divide the length of the string by the circumference.

Step 1) Find the circumference

C = pi x diameter

C = 3.14 x 3

C = 9.42

Step 2) Divide the length of the string by the circumference

90.42 / 9.42 = 9.5987

The string would make approximately 9.60 turns around the test tube.

Hope this helps!! :)

6 0

3 years ago

What set of transformations is performed on triangle ABC to form triangle A′B′C′? (4 points)

fenix001 [56]

Answer:

i am on this exact question and i am confused myself because it seems like the answer should be "180 degree rotation with a translation of 5 units to the right" but that is not an option :/ so my best guess is either A or D hope this helps at all

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Can someone plz help me. How can you find the inequalities of 11/15 and 5/7. Next 5/9 and 7/13. Next 11/15 and 5/7. Lastly 5/9 a

LekaFEV [45]

To make this a little clearer, let's give the pairs of inequalities the same denominator:

Question 1:
 $\frac{11}{15}$ ? $\frac{5}{7}$
First, apply the common denominator to the first fraction:
$(\frac{11}{15})7 \\ \frac{11}{15} * \frac{7}{7} \\ \frac{11*7}{15*7} \\ \frac{77}{105}$
Do the same for the second: $15( \frac{5}{7}) \\ \frac{5}{7}* \frac{15}{15} \\ \frac{5*15}{7*15} \\ \frac{75}{105}$
Nest, compare the two fractions:
$\frac{77}{105} \ \textgreater \ \frac{75}{105}$
Therefore:
$\frac{11}{15}$ > $\frac{5}{7}$

Question Two:
$\frac{5}{9}$ ? $\frac{7}{13}$
Apply the common denominator to fraction one:
$13( \frac{5}{9}) \\ \frac{5}{9} * \frac{13}{13} \\ \frac{5*13}{9*13} \\ \frac{65}{117}$
Fraction two:
$9(\frac{7}{13}) \\ \frac{7}{13} * \frac{9}{9} \\ \frac{7*9}{13*9} \\ \frac{63}{117}$
Evaluate:
$\frac{65}{117}$ > $\frac{63}{117}$
Therefore:
 $\frac{5}{9}$ > $\frac{7}{13}$

Hope this helps!

5 0

3 years ago