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4 years ago

Question 1 of 10

Mathematics

1 answer:

AleksandrR [38]4 years ago

7 0

Answer:

Step-by-step explanation:

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If a = 3, what is the value of 2a2. A. 6 B. 12 C. 18 D. 36

aivan3 [116]

Answer:

Step-by-step explanation:

I got it correct good luck bye!

6 0

3 years ago

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Trevor spent $27 and now has no money left. What did he have before his purchase

Sav [38]

Answer:$27

Step-by-step explanation:

x-27=0

0+27=27

27-27=0

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3 years ago

Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

we can simplify left side and make it equal to right side

we can use trig identity

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

$\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}$

$\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}$

$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

$\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can use trig identity

$sin^2(a)+cos^2(a)=1$

$\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can cancel terms

$=3-4(1-sin(a)cos(a))$

now, we can simplify it further

$=3-4+4sin(a)cos(a))$

$=-1+4sin(a)cos(a))$

$=4sin(a)cos(a)-1$

$=2\times 2sin(a)cos(a)-1$

now, we can use trig identity

$2sin(a)cos(a)=sin(2a)$

we can replace it

$=2sin(2a)-1$

so,

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

7 0

3 years ago

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Please tell me what x is 3(x+6)=48

Alekssandra [29.7K]

Answer

X=10

Step by step explanation

48÷3 = 16

16-6 = 10

7 0

4 years ago

What are solutions to the equation below x^2 +3x-18=0

Ivenika [448]

$x^2 +3x-18=0\\ x^2-3x+6x-18=0\\ x(x-3)+6(x-3)=0\\ (x+6)(x-3)=0\\ x=-6 \vee x=3$

5 0

3 years ago

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