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Nezavi [6.7K]
4 years ago
5

Can some body help me with this geometry question please I need the answer asap

Mathematics
1 answer:
velikii [3]4 years ago
4 0
A.)  m∠EBD = m∠EBC - m∠CBD = 140 - 35 = 105

      m∠ABE = 180 - m∠EBC = 180 - 140 = 40


b.)  no because the angle measure is not 90 degrees
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Suppose we have a right triangle with legs of length a and b and hypotenuse of length c. Suppose b=3 and c=5. Then a= , For the
ANTONII [103]

Answer:

Length of right-angle  triangle 'a' = 4

b)

<u><em></em></u>sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}<u><em></em></u>

<u><em></em></u>cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}<u><em></em></u>

<u><em></em></u>tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}<u><em></em></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given  b = 3 and hypotenuse c = 5

Given ΔABC  is a right angle triangle

By using pythagoras theorem

        c² = a² + b²

  ⇒ a² = c² - b²

 ⇒  a² = 5²-3²

          =25 - 9

      a² = 16

⇒   a = √16 = 4

The sides of right angle triangle  a = 4 ,b = 3 and c = 5

<u><em>Step(ii):-</em></u>

<u><em></em></u>sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}<u><em></em></u>

<u><em></em></u>cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}<u><em></em></u>

<u><em></em></u>tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}<u><em></em></u>

7 0
3 years ago
I will mark the brainiest! Please, and thanks! :)
serious [3.7K]

Answer:

I added the picture as an attached file below

Step-by-step explanation:

Hope it helps:)

6 0
4 years ago
X + 4 8. If you need to solve for x, what's the first step?
natta225 [31]

Answer:

B) Subtract 4 from both sides

Step-by-step explanation:

You do this so you can get x alone, which then shows you what the answer is

x=4

8 0
4 years ago
How do I do question 9? Please give answer thank you
kvasek [131]
<h3>Given</h3>
  • a cone of height 0.4 m and diameter 0.3 m
  • filling at the rate 0.004 m³/s
  • fill height of 0.2 m at the time of interest
<h3>Find</h3>
  • the rate of change of fill height at the time of interest
<h3>Solution</h3>

The cone is filled to half its depth at the time of interest, so the surface area of the filled portion will be (1/2)² times the surface area of the top of the cone. The filled portion has an area of

... A = (1/4)(π/4)d² = (π/16)(0.3 m)² = 0.09π/16 m²

This area multiplied by the rate of change of fill height (dh/dt) will give the rate of change of volume.

... (0.09π/16 m²)×dh/dt = dV/dt = 0.004 m³/s

Dividing by the coefficient of dh/dt, we get

... dh/dt = 0.004·16/(0.09π) m/s

... dh/dt = 32/(45π) m/s ≈ 0.22635 m/s

_____

You can also write an equation for the filled volume in terms of the filled height, then differentiate and solve for dh/dt. When you do, you find the relation between rates of change of height and area are as described above. We have taken a "shortcut" based on the knowledge gained from solving it this way. (No arithmetic operations are saved. We only avoid the process of taking the derivative.)

Note that the cone dimensions mean the radius is 3/8 of the height.

V = (1/3)πr²h = (1/3)π(3/8·h)²·h = 3π/64·h³

dV/dt = 9π/64·h²·dh/dt

.004 = 9π/64·0.2²·dh/dt . . . substitute the given values

dh/dt = .004·64/(.04·9·π) = 32/(45π)

7 0
3 years ago
What is the equation that could be used to solve for x?
pashok25 [27]

Answer:

x=32

Step-by-step explanation:

I believe the angles are supplementary so you could use the equation:

2x+8 + 3x+22 = 180

5x+30=180

5x=160

x=32

6 0
3 years ago
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