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geniusboy [140]
3 years ago
5

What is the area of the trapezoid below? 10 ft/12 ft/13ft/14ft

Mathematics
2 answers:
creativ13 [48]3 years ago
8 0
Area of trapezoid  = 1/2 (sum of parallel sides) × (perpendicular distance between them)

Area = 1/2 × (3 + 7) × 2 =  10 ft²
Pavlova-9 [17]3 years ago
3 0
The answer is definitely 10ftsquared
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Find sin(2x), cos(2x), and tan(2x) from the given information.<br> tan(x) = − 1/2 , cos(x) &gt; 0
vodomira [7]

Answer:

  • sin(2x) = -4/5
  • cos(2x) = 3/5
  • tan(2x) = -4/3

Step-by-step explanation:

It may be easiest to start with tan(2x).

  tan(2x) = 2tan(x)/(1 -tan(x)²)

  tan(2x) = 2(-1/2)/(1 -(-1/2)²) = -1/(3/4)

  tan(2x) = -4/3 . . . . . still a 4th-quadrant angle

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Then cosine can be found from ...

  cos(2x) = 1/√(tan(2x)² +1) = 1/√((-4/3)²+1) = √(9/25)

  cos(2x) = 3/5

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Sine can be found from these two:

  sin(2x) = cos(2x)tan(2x) = (3/5)(-4/3)

  sin(2x) = -4/5

5 0
3 years ago
How would you explain this methodology and could you identify the independent, dependent, control variables, how would you defen
musickatia [10]

Christmas font Courier South Crystal Downs matemáticas

3 0
3 years ago
There are four roads from A to B and five roads from B to C. In how mamy different ways can a person travel from A to C?
Lelu [443]

That would be 4 * 5 = 20 ways  answer.

3 0
3 years ago
Read 2 more answers
BAC EDF BAC is 24 what is EDF
allochka39001 [22]

Answer:

6inches^2

Step-by-step explanation:

7 0
3 years ago
How do you do this question?
Sergeeva-Olga [200]

Answer:

V = 2000r³/3

Step-by-step explanation:

We know that the base is a circular disk, so it creates a circle on the xy plane. It would be in the form x² + y² = r². In other words x² + y² = (5r)². Let's isolate y in this equation now:

x² + y² = (5r)²,

x² + y² = 25r²,

y² = 25r² - x²,

y = √25r² - x² ---- (1)

Now remember that parallel cross sections perpendicular to the base are squares. Therefore Area = length^2. The length will then be = 2√25r² - x² --- (2). Now we can evaluate the integral from -5r to 5r, of [ 2√25r² - x² ]² dx.

\int _{-5r}^{5r}\:\left[\:2\sqrt{\left(25r^2\:-\:x^2\right)}\:\right]\:^2\:dx\\=\int _{-5r}^{5r}4\left(25r^2-x^2\right)dx\\\\= 4\cdot \int _{-5r}^{5r}25r^2-x^2dx\\\\= 4\left(\int _{-5r}^{5r}25r^2dx-\int _{-5r}^{5r}x^2dx\right)\\\\= 4\left(250r^3-\frac{250r^3}{3}\right)\\\\= 4\cdot \frac{500r^3}{3}\\\\= \frac{2000r^3}{3}

As you can see, your exact solution would be, V = 2000r³/3. Hope that helps!

3 0
2 years ago
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