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TiliK225 [7]
3 years ago
10

cos (x/2) = -sqrt2/2 in (0,360) looking for two values ?? Can someone please help me I can’t seem to figure out how to solve thi

s problem, I’m looking at other examples but I’m still confused please help
Mathematics
1 answer:
ValentinkaMS [17]3 years ago
8 0

Answer:

135° and 225°

Step-by-step explanation:

basically you want to find the value of x between 0 and 360 in this equation

cos x/2 = -(√2)/2

assume x/2 as n, so

cos n = -(√2)/2

n = 45°

then remember the quadrant system

0-90 1st quadrant, all is POSITIVE

90-180 2nd quadrant, only SIN has positive value

180 - 270 3rd quadrant, only TAN has positive value

270 -360 4th quadrant, only COS positive here.

so if you try to find negative value look into 2nd and 3rd quadrant that related 45° to x-axis (0°or 180°)

so the value of x is

180 - 45 = 135° (2nd quadrant) and

180 + 45° = 225° (3rd quadrant)

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The school that Mary goes to is selling tickets to a spring musical. On the first day of ticket sales
andrezito [222]

Answer:

The cost of one adult ticket is $13, and the price of one student ticket is $4.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

x is the cost of an adult ticket

y is the cost of a student ticket.

6 adult tickets and 1 student ticket for a total of $82

This means that

6x + y = 82

y = 82 - 6x

The school took in $51 on the second day by selling 3 adult tickets and 3 student tickets.

This means that

3x + 3y = 51

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x + y = 17

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x + 82 - 6x = 17

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4 0
3 years ago
Please help me again
gulaghasi [49]

Answer:

65 degrees

Step-by-step explanation:

<u>Step 1:  Find what the measure of JLK is</u>

JLM + JLK = 180

140 + JLK - 140 = 180 - 140

JLK = 40

<u>Step 2:  Find what the measure of KJL is</u>

KJL + JLK + JKL = 180

KJL + 40 + 75 = 180

KJL + 115 - 115 = 180 - 115

KJL = 65

Answer:  65 degrees

5 0
3 years ago
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