There are 2 choices for the first set, and 5 choices for the second set. Each of the 2 choices from the first set can be combined with each of the 5 choices from the second set. Therefore there are 2 times 5 combinations from the first and second sets. Continuing this reasoning, the total number of unique combinations of one object from each set is:
I believe it is d also hope that help
Answer:
the forth one is linear
Step-by-step explanation:
That's a trapezoid,
Answer: 32
But say we didn't know the formula. We can chop out the 4 by 4 square and we have two 45/45/90 triangles side 4 left over, so two 4 by 4 squares,
A = 2(4)(4) = 32
We can think of this as 12 by 4 rectangle with two 4 by 4 right triangles chopped out:
A = 12(4) - 2(1/2)(4)(4) = 48 - 16 = 32
The exact value is
<span>sin<span>(arccos<span>(<span>3/4</span>)</span>)</span></span>The equation for cosine is <span>cos<span>(A)</span>=<span>AdjacentHypotenuse</span></span>. The inside trig function is <span>arccos<span>(<span>3/4</span>)</span></span>, which means <span>cos<span>(A)</span>=<span>3/4</span></span>. Comparing <span>cos<span>(A)</span>=<span>AdjacentHypotenuse</span></span> with <span>cos<span>(A)</span>=<span>3/4</span></span>, find <span>Adjacent=3</span> and <span>Hypotenuse=4</span>. Then, using the pythagorean theorem, find <span>Opposite=<span>√7</span></span>.<span>Adjacent=3</span><span>Opposite=<span>√7</span></span><span>Hypotenuse=4</span>Substitute in the known variables for the equation <span>sin<span>(A)</span>=<span>OppositeHypotenuse</span></span>.<span>sin<span>(A)</span>=<span><span>√7</span> over 4</span></span>Simplify.<span><span>√7</span><span> over 4</span></span>
