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Andreas93 [3]
3 years ago
14

At a food festival, 3/8 of the dishes were from china. Another 12.5% of the dishes were from japan. What percent of the dishes w

ere from the other countries?
Mathematics
1 answer:
MrMuchimi3 years ago
5 0

Answer: 50%

Step-by-step explanation:

Let x = Total dishes.

Dishes from China = \dfrac38x

Dishes from Japan  = 12.5\% \text{ of }x= 0.125x

Total dishes  from China and Japan = \dfrac38 x+0.125x=0.375x+0.125x=0.5x

Dishes from other countries = x- 0.5x= 0.5x

Percent of dishes from other countries= \dfrac{0.5x}{x}\times100\%= 50\%

Hence, dishes from other countries = 50%

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gavmur [86]

Answer:

5. DBE and CBA both=48 degrees

6. a. scalene, right

b.isosceles, obtuse

c. scalene, acute

7. 50 degrees

8. im not completely sure how to do this one but ill check into it

Step-by-step explanation:

its a lot to explain, in 5 find p and plug it in.

in 6, just look at the sides and angles.

scalene- no sides equal

isosceles- 2 sides equal

equilateral- all sides equal

right- one 90 degree angle

acute- all acute angles (less than 90)

obtuse- one obtuse angle (more than 90)

7. add 90 and 40 to get 130. 180-130=50

8. im gonna look up how to do this one right now

6 0
3 years ago
Lowering powers write in terms of first power of cosine. Cos^6
yaroslaw [1]

The main identity you need is the double angle one for cosine:

\cos^2x=\dfrac{1+\cos2x}2

We get

\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8

Expand the numerator to apply the identity again:

\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8

\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8

\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x

Finally, make use of the product identity for cosine:

\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2

so that ultimately,

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x

\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x

4 0
3 years ago
in the figure below , Line K is perpendicular to Line l. based on the information given in the figure, find the measure of <C
gulaghasi [49]
C is the answer for this
8 0
3 years ago
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In ΔCDE, m∠C = (9x+11)^{\circ}(9x+11) ∘ , m∠D = (3x-15)^{\circ}(3x−15) ∘ , and m∠E = (2x+16)^{\circ}(2x+16) ∘ . What is the valu
Ksivusya [100]

Answer:

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Step-by-step explanation:

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Sum the 3 given angles and equate to 180, that is

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14x + 12 = 180 ( subtract 12 from both sides )

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5 0
3 years ago
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nataly862011 [7]

Answer:

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Step-by-step explanation:

x=5

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2 years ago
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