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s344n2d4d5 [400]
4 years ago
9

Solve the equation 2032 * 93 equals?

Mathematics
1 answer:
AleksAgata [21]4 years ago
3 0

Answer:

188976

Step-by-step explanation:

2032 x 93 = 188976

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Prove the converse of Thales’ theorem: If AB is a diameter of a circle and P is a point so that ∠????PB is a right angle,
olasank [31]

Idk but if anyone does know it because im stuck too!

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3 years ago
Hi, I am very stuck on questions 17 and 18. Can someone help please? Thanks
NISA [10]

9514 1404 393

Answer:

  17) x = 10√2

  18) x = 4√6

Step-by-step explanation:

These problems rely on your knowledge of the side ratios of special triangles

  45°-45°-90° triangle: 1 : 1 : √2

  30°-60°-90° triangle: 1 : √3 : 2

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17) 5 is the short side of the isosceles right triangle, so its hypotenuse is 5√2. That length is the shortest side of the 30/60/90 triangle, so its longest side is 2 times that:

  x = 10√2

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18) 8 is the longest side of the 30/60/90 triangle, so its long leg will be 8(√3)/2 = 4√3. The hypotenuse of the isosceles right triangle is √2 times that, so ...

  x = (4√3)(√2)

  x = 4√6

6 0
3 years ago
What is f(x)=square root of 2x g(x)=square root of 32 x. Find (f timesheet)(x)
Alexus [3.1K]
B is the correct answer
6 0
3 years ago
Someone please help me
Free_Kalibri [48]
D. y = 3x + 2

Check (-4, -10)
-10 = 3(-4) + 2
-10 = -12 + 2
-10 = -10 :)

Check (-3, -7)
-7 = 3(-3) + 2
-7 = -9 + 2
-7 = -7 :)

Check (-2, -4)
-4 = 3(-2) + 2
-4 = -6 + 2
-4 = -4 :)

Check (-1, -1)
-1 = 3(-1) + 2
-1 = -3 + 2
-1 = -1 :)

Check (0, 2)
2 = 3(0) + 2
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4 0
3 years ago
A sine function had an amplitude of 3, period of 6pi, horizontal shift of 3pi/2, & vertical shift of -1.
Simora [160]

Answer: \bold{y=\dfrac{1}{2}}

<u>Step-by-step explanation:</u>

f(x) = A sin (Bx - C) + D

  • amplitude = |A|
  • period =\dfrac{2\pi}{B}
  • phase shift =\dfrac{C}{B}
  • vertical shift = D

<u>A</u>

amplitude of 3 is given so  3 = |A| → A = ± 3, since it is stated that this is a positive function, then A = 3

<u>B</u>

period of 6π is given so 6\pi=\dfrac{2\pi}{B}\quad \rightarrow \quad B=\dfrac{2\pi}{6\pi}\quad \rightarrow \quad B=\dfrac{1}{3}

<u>C</u>

\text{phase shift is given as}\ \dfrac{3\pi}{2}\ \text{so}\ \dfrac{3\pi}{2}=\dfrac{C}{\frac{1}{3}}\quad \rightarrow\quad \dfrac{(\frac{1}{3})3\pi}{2}=C\quad \rightarrow\quad \dfrac{\pi}{2}=C

<u>D</u>

vertical shift of -1 is given so -1 = D


Now, substitute the values of A, B, C, and D into the formula (above):

f(x) = 3\ sin \bigg(\dfrac{1}{3}x - \dfrac{\pi}{2}\bigg) - 1


Next, solve when x = 2π

f(2\pi) = 3\ sin \bigg(\dfrac{1}{3}(2\pi) - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\bigg) - 1

        = 3\ sin \bigg(\dfrac{4\pi}{6} - \dfrac{3\pi}{6}\bigg) - 1

        = 3\ sin \bigg(\dfrac{\pi}{6}\bigg) - 1

        = 3\ \bigg(\dfrac{1}{2}\bigg) - 1

        =\dfrac{3}{2}-\dfrac{2}{2}

        =\dfrac{1}{2}

6 0
3 years ago
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