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kogti [31]
3 years ago
11

Solve for x 15cot(2x)+3=0 please help

Mathematics
1 answer:
goldenfox [79]3 years ago
5 0
<span>The best answer   Is
the values of x are -0.6867and 0.8841</span>
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10% of £26 is equal to x% of £52. Calculate the value of x.<br><br>​
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5

Step-by-step explanation:

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x = 5

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Alison wants to find out how much time people spend reading books. She is going to use a questionnaire.
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5 0
3 years ago
If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num
weeeeeb [17]

Answer:

<h2>x = 0, y = 5, z = 3</h2>

Step-by-step explanation:

x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right

\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right

\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}

7 0
3 years ago
Given: ABCD is a trapezoid, AB = CD, BK ⊥ AD, AK = 10, KD = 20 Find: BC AD
alexdok [17]

Answer:

<u> BC = 10 and AD = 30</u>

Step-by-step explanation:

In figure-1 , AB = CD ,BK ⊥ AD,  AK = 10,  KD = 20.

Since, line AD is sum of AK and  KD, then

AD = AK + KD

AD = 10 + 20

AD = 30

Since, BC ║AD and BK ⊥ AD then similarly we construct CL ⊥ AD

so, BC = KL and AK = LD

KL = AD - LD

KL = 20 - 10

KL = 10

Since, BC = KL then BC = 10

Hence, <u> BC = 10 and AD = 30</u>

4 0
3 years ago
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