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Darina [25.2K]
3 years ago
5

Find the GCF of 75, 100, and 175.

Mathematics
2 answers:
nataly862011 [7]3 years ago
6 0
It's twenty-five, it's simple, try khan academy if your really struggling


Julli [10]3 years ago
3 0
25
 I think I'm not completely sure
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Find the tangents of the acute angles in the right triangle. Write each answer as a fraction. WILL MAKE BRAINLIEST!!
patriot [66]

Answers:

  • tan(G) = 2
  • tan(H) = 1/2

=======================================

Explanation:

Recall that tangent is the ratio of opposite over adjacent

tan(angle) = opposite/adjacent

So for reference angle G, we say,

tan(G) = JH/GJ = 2/1 = 2

We'll treat tan(H) in a similar fashion, but the opposite and adjacent sides swap roles. That means we'll apply the reciprocal to the result above to get 1/2 for tan(H)

-----------

So we have this interesting property where

tan(G)*tan(H) = 2*(1/2) = 1

In general,

tan(A)*tan(B) = 1 if and only if A+B = 90 degrees

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Side note: The side sqrt(5) isn't used at all.

8 0
3 years ago
Combine the following expressions.
Keith_Richards [23]

Answer:

second option

Step-by-step explanation:

Using the rule of radicals

\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

Simplifying the radicals

\sqrt{27x^{3} }

= \sqrt{9(3)x^2.x}

= \sqrt{9} × \sqrt{3} × \sqrt{x^2} × \sqrt{x}

= 3 × \sqrt{3} × x × \sqrt{x}

= 3x\sqrt{3x}

-------------------------------

\sqrt{12x^{3} }

= \sqrt{4(3)x^2.x}

= \sqrt{4} × \sqrt{3} × \sqrt{x^2} × \sqrt{x}

= 2 × \sqrt{3} × x × \sqrt{x}

= 2x\sqrt{3x}

---------------------------------------

Thus

3 × 3x\sqrt{3x} - 2 × 2x\sqrt{3x}

= 9x\sqrt{3x} - 4x\sqrt{3x}

= 5x\sqrt{3x}

8 0
3 years ago
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