Can someone please help me on this geometry test, serious inquiries only, be a real one

Mathematics

1 answer:

saveliy_v [14]3 years ago

5 0

I took geometry a while ago, I can help if you want.

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Simplify by dividing 3/5 ÷ -4/9 1. -27/20 2. 27/20 3. -12/45 4. 12/45

kramer

Answer:

The answer to your question is -27/20

Step-by-step explanation:

Divide 3/5 ÷ - 4/9

Process

1.- Just multiply the numerator of the first fraction by the denominator of the second fraction.

3 x - 9 = -27

2.- Multiply the denominator of the first fraction by the numerator of the second fraction.

5 x 4 = 20

3.- Join both results

-27/20

5 0

3 years ago

I need help if you get it right I will mark you as brainlest

musickatia [10]

Answer:

D. 9/DF=18/16.......the answer

3 0

3 years ago

I am offering another 100 points

Ivahew [28]

Answer:

$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

Step-by-step explanation:

Perfect squares: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...

To find $\sf \sqrt{75}$ , identify the perfect squares immediately before and after 75:

64 and 81

$\begin{aligned}\sf As\;\; 64 < 75 < 81\; & \implies \sf \sqrt{64} < \sqrt{75} < \sqrt{81}\\&\implies \sf \;\;\;\;\;8 < \sqrt{75} < 9 \end{aligned}$

$\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.$

See the attachment for the correct placement of $\sf \sqrt{75}$ on the number line.