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frozen [14]
3 years ago
15

Use the unit circle to find the value of sin(3pi/2) and cos(3pi/2). hurry I need help ASAP!

Mathematics
1 answer:
Marta_Voda [28]3 years ago
4 0

Answer:

Step-by-step explanation:

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The diagonal of a rectangle is of length a. It splits each corner forming two angles with a ratio of 1:2. The area of the rectan
HACTEHA [7]

Answer:

Step-by-step explanation:

Given

Length of diagonal is a

Diagonal divides the angle in 1:2

such that \theta +2\theta =90 (because angle between two sides is 90)

3\theta =90

\theta =30^{\circ}

width of rectangle is b=a\sin \theta =\frac{a}{2}

Length of rectangle is L=a\cos 30=\frac{\sqrt{3}}{2}a

Area of rectangle A=L\cdot b

A=\frac{\sqrt{3}}{2}a\times \frac{a}{2}

A=\frac{\sqrt{3}}{4}a^2

5 0
3 years ago
Use quadratic formula to solve 1-6x^2-x​
eduard

Answer:

(3x-1)(2x+1)

Step-by-step explanation:

1-6x^2-x=0

-6x^2-x+1=0

6x^2+x-1=0

6x^2+(3-2)x-1=0

6x^2+3x-2x-1=0

3x(2x+1)-1(2x+1)=0

(3x-1)(2x+1)=0

So the factor is (3x-1)(2x+1)

6 0
3 years ago
Double a number decreased by 4
Wittaler [7]

Answer:

2x-5 if u want u can substitute any other variables and constant

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Y=2(x+1)^2 has how many real roots
oee [108]

The equation y= 2(x+1)^2 has one real root and that is x=-1.

What is real roots of the equation?

    We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation X^2-7x+12=0 is solved, the actual roots are 3 and 4.

Here given,

=> y = 2(x+1)^2

Take y=0 then,

=> 2(x+1)^2=0

=> (x+1)^2=0

=>(x+1)=0

=> x=-1

Hence the given equation has one real root and that is x=-1.

To learn more about real roots refer the below link

brainly.com/question/24147137

#SPJ1

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1 year ago
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Can someone help me find the domain of this function please ?
ratelena [41]

Answer: It is A=-1<x<-6 is the domain

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3 years ago
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