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hjlf
4 years ago
8

How do I know that the following equation is true:

Mathematics
1 answer:
PSYCHO15rus [73]4 years ago
4 0

Answer:

<u>Half angle Identity for cosine</u>:

$\cos \left(\frac{\theta}{2} \right)=\pm \sqrt{\frac{1+\cos(\theta)}{2} } $

\text{Note that it is} \pm \text{because we don't know which Quadrant the half-angle is, }\text{the information of}\cos(\theta) \text{is not enough to conclude the angle}

$\frac{\pi}{8} \text{ is a Quadrant I angle, therefore, positive}$

$\cos \left(\frac{\pi}{8} \right)=\sqrt{\frac{1+\cos(\frac{\pi}{4} )}{2} } $

The equation is true.

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B is the answer !!!!!
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3 years ago
Joanne had $115 to spend at the mall. She spent 23% of that money on a shirt. How much did Joanne spend on the shirt?
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The answer would be $26.45.
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3 years ago
The length of a rectangle is five centimeters less than width. Find the dimensions of the
kondaur [170]

Answer:

w is 10 and l is 5

Step-by-step explanation:

l=w-5

lw=50

w^2-5w-50=(w-10)(w+5) since the negative solution does not make sense w=10 and therefore l is 5 so the length is 5 while the width is 10

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3 years ago
Please help me answer this i need asap
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Answer:24 xexponet 3 − 5

Step-by-step explanation:

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3 years ago
Find the intersection points using substitution or elimination for each system of equations:
natka813 [3]

Answer:

Step-by-step explanation:

Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.

Problem 1:

(x²)/4 +y²= 1

y= x+1

*substitute for y*

Now we have a one-variable equation we can solve-

x²/4 + (x+1)² = 1

x²/4 + (x+1)(x+1)= 1

x²/4 + x²+2x+1= 1

*subtract 1 from both sides to set equal to 0*

x²/4 +x^2+2x=0

x²/4 can also be 1/4 * x²

1/4 * x² +1*x² +2x = 0

*combine like terms*

5/4 * x^2+2x+ 0 =0

now, you can use the quadratic equation to solve for x

a= 5/4

b= 2

c=0

the syntax on this will be rough, but I'll do my best...

x= (-b ± √(b²-4ac))/(2a)

x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))

x= (-2 ±√(4-0))/(2.5)

x= (-2±2)/2.5

x will have 2 answers because of ±

x= 0 or x= 1.6

now plug that back into one of the equations and solve.

y= 0+1 = 1

y= 1.6+1= 2.6

Hopefully this explanation was enough to help you solve problem 2.

Problem 2:

x² + y² -16y +39= 0

y²- x² -9= 0

6 0
3 years ago
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