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Vedmedyk [2.9K]
2 years ago
12

What would be an equation for this

Mathematics
2 answers:
wolverine [178]2 years ago
6 0

Answer: 19 - 4 = 7 + x

guajiro [1.7K]2 years ago
4 0
Answer:

4x-19=1/2x+7
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Sin x+sin 5x= cot 2x​
butalik [34]

Step-by-step explanation:

  1. sinx+sin(x+4x) =cot2x
  2. sinx+sinx cos4x+cosx sin4x=cot2x

sinx+sinx cos4x=sinx (1+cos4x) =

sinx (1-1+2(cos2x)^2) =2sinx (cos2x)^2

sin4x=2sin2xcos2x

2sinx(cos2x)^2+cosx+2sin2xcos2x=cot2x

2sinxcos2x(cos2x+2(cosx)^2)=cot2x

but 2(cosx)^2=cos2x+1

2sinxcos2x(cos2x+cos2x+1)=cos2x/sin2x

(2sinxsin2x)(2cos2x+1) =1

(4cosx(sinx)^2)(2cos2x+1) =1

6 0
2 years ago
Solve for x: x/3 = -48
oksano4ka [1.4K]

Answer:

-144

Step-by-step explanation:

x=-48 times 3

8 0
2 years ago
Read 2 more answers
Show the working out
m_a_m_a [10]

Answer:

59.25 currency ; No. of cans rounded to next whole number

Step-by-step explanation:

Draw a perpendicular line frrom the 12m. line onto the 20m. line. We get a rectangle on the left and a right-triangle on the right.

Using Pythagorous theorem,

Fence = \sqrt{17^{2} - 8^{2}} m = 15 m

So, area of the lawn = ( 1/2 ) * ( 12 + 20 ) * 16 sqm = 240 sqm

So, number of cans needed = 240/100 = 2.4 ≈ 3 [Cause cans can't be partially bought ]

Total cost = 3 * 19.75 currency = 59.25 currency

( However, if cans can be partially bought, then total cost = 47.4 currency )

4 0
3 years ago
A caterer charger $120 to carter a party for 15 people and $200 for 25 people
Luba_88 [7]

Answer:

y = 8x

Step-by-step explanation:

When the number of people increased by 10, the caterer charged an additional $80. Therefore, we estimate that the cost per person is $8. In order to verify:

$8/person * 15 people = $120. True. $8/person * 25 people = $200. True.

Therefore, if we let x represent the number of people being cater for and y represent the cost, the equation to model the situation is simply y = 8x.

hope it helps!

5 0
2 years ago
There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student
Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0
3 years ago
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