Fourteen eleventh graders prefer math and eight tenth graders prefer English.
<h3>What is a survey?</h3>
A survey is a study that seeks to gather information about people sometimes involving the preferences of the people.
From the table, we can see that fourteen eleventh graders prefer math and eight tenth graders prefer English.
Learn more about survey:brainly.com/question/3944932
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Answer:
105degrees
Step-by-step explanation:
Using the theorems that states that the sum of the opposite side of the quadrilateral is 180degrees, hence;
<RST + <TQR = 180
5x+15 + 4x+ 3 = 180
9x + 18 = 180
9x = 180 - 18
9x = 162
x = 162/9
x = 18
Since <RST = 5x+15
<RST = 5(18) + 15
<RST = 90 + 15
<RST = 105degrees
Hence the measure of <RST is 105degrees
Answer:
Therefore the height of the tower is 116.75 m.
Step-by-step explanation:
Given that a pole that is 2.6 m tall casts a shadow that is 1.03 m long.
Here the height of a object directly proportional to the shadow. It means
Height of the object ∝ shadow
Height of the object= k.shadow
[k is the proportional constant.]
[Height of object is denoted by h and shadow is denoted by s]
Then,
[h₁ is the height of the pole and s₁is the length of the shadow]
Again for the tower
[h₂ is the height of the tower and s₂ is the length of the shadow]
Therefore,
![\frac{h_1}{s_1}=\frac{h_2}{s_2} =k](https://tex.z-dn.net/?f=%5Cfrac%7Bh_1%7D%7Bs_1%7D%3D%5Cfrac%7Bh_2%7D%7Bs_2%7D%20%3Dk)
............(1)
Given
,
and
Putting the value of
and
equation (1)
![\frac{2.6}{1.03}=\frac{h_2}{46.25}](https://tex.z-dn.net/?f=%5Cfrac%7B2.6%7D%7B1.03%7D%3D%5Cfrac%7Bh_2%7D%7B46.25%7D)
[cross multiplication]
![\Rightarrow h_2 = \frac{2.6\times 46.25}{1.03}](https://tex.z-dn.net/?f=%5CRightarrow%20h_2%20%3D%20%5Cfrac%7B2.6%5Ctimes%2046.25%7D%7B1.03%7D)
![\Rightarrow h_2= 116.75](https://tex.z-dn.net/?f=%5CRightarrow%20h_2%3D%20116.75)
Therefore the height of the tower is 116.75 m.
Answer:
1.23, 0.067, and 25.34
Step-by-step explanation:
Hope this helps!
brainliest?
:)
∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.
Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².
∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find
![\sin(30^\circ) = \dfrac{h}{4\,\rm cm} \implies h= 2\,\rm cm](https://tex.z-dn.net/?f=%5Csin%2830%5E%5Ccirc%29%20%3D%20%5Cdfrac%7Bh%7D%7B4%5C%2C%5Crm%20cm%7D%20%5Cimplies%20h%3D%202%5C%2C%5Crm%20cm)
where
is the length of the altitude originating from vertex O, and so
![\left(\dfrac b2\right)^2 + h^2 = (4\,\mathrm{cm})^2 \implies b = 4\sqrt3 \,\rm cm](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%20b2%5Cright%29%5E2%20%2B%20h%5E2%20%3D%20%284%5C%2C%5Cmathrm%7Bcm%7D%29%5E2%20%5Cimplies%20b%20%3D%204%5Csqrt3%20%5C%2C%5Crm%20cm)
where
is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².
So, the total area of the shaded region is
(8π/3 - 4√3) + (16π/3 - 4√3) = (8π - 8√3) cm²