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RSB [31]
3 years ago
7

The semicircle shown at left has center X XX and diameter W Z ‾ ​WZ ​ ​​ start overline, W, Z, end overline. The radius X Y ‾ ​X

Y ​ ​​ start overline, X, Y, end overline of the semicircle has length 2 22. The chord Y Z ‾ ​YZ ​ ​​ start overline, Y, Z, end overline has length 2 22. What is the area of the shaded sector formed by obtuse angle W X Y WXYW, X, Y?
Mathematics
1 answer:
BabaBlast [244]3 years ago
7 0
Triangle XYZ is equilateral, so the central angle of the sector is 120°, or 2π/3 radians. The area of a sector of central angle β is given by
.. A = (1/2)r^2*β . . . . . . β in radians
Your sector's area is
.. A = (1/2)*2^2*(2π/3) = 4π/3 . . . . square units
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Together, angles A and B form a 90° angle. Angle A measures 54°. What is the measure of angle B?
Hunter-Best [27]

Answer:

36 degrees

Step-by-step explanation:

Subtract 54 from 90

7 0
3 years ago
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Find the volume and area for the objects shown and answer Question
klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of ​​the given solids.

\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2

\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\

\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}

\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}

SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}

6 0
3 years ago
Urgent please help..
Alexxx [7]

Answer:

x ≈ 13.5

Step-by-step explanation:

Using the sine rule in the triangle, then

\frac{sin34}{x} = \frac{sin27}{11} ( cross- multiply )

x × sin27° = 11 × sin34° ( divide both sides by sin27° )

x = \frac{11sin34}{sin27} ≈ 13.5 ( to the nearest tenth )

5 0
2 years ago
Log base 2 (1/4) ...?
liubo4ka [24]
Log ( base 2 ) ( 1 / 4 ) =
= log ( base 2 ) ( 2 ^(-2 ) ) =   
= - 2 log ( base 2 ) 2 =    ( because : log x^n = n log x )
= ( - 2 ) * 1 =    ( because:   log (base x) x = 1 ) 
= -  2
5 0
3 years ago
What is the equation of the line that passes through (-3,-1) and has a slope of 2/5?
barxatty [35]

Answer:

y = mx + b

-1 = 2/5(-3) + b

-1 = -6/5+b

-1 + 6/5 = -6/5 + 6/5 + b

1/5 = b

y = 2/5x + 1/5

Step-by-step explanation:

  1. write down the slope-intercept form formula
  2. identify your x and y of the point
  3. replace x and y by the point
  4. multiply m and x
  5. do addition to get b
  6. write down the equation by replacing m and b with their values
3 0
3 years ago
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