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3 years ago

7-8÷4(3^2-1) ~answer step by step please~

Mathematics

1 answer:

Schach [20]3 years ago

5 0

Answer:

-9

Step-by-step explanation:

Simplify the following:

7 - 8/4 (3^2 - 1)

Hint: | Evaluate 3^2.

3^2 = 9:

7 - 8/4 (9 - 1)

Hint: | Subtract 1 from 9.

9 - 1 = 8:

7 - 2×8

Hint: | Divide 8 by 4.

8/4 = (4×2)/4 = 2:

7 - 2×8

Hint: | Multiply -2 and 8 together.

-2×8 = -16:

7 + -16

Hint: | Subtract 16 from 7.

7 - 16 = -9:

Answer: -9

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Determine the center and radius of the following circle equation:

UkoKoshka [18]

Answer:

The equation of the circle (x + 5 )² + ( y + 10 )² = (4)²

Center of the circle ( h, k) = ( -5 , -10)

Radius of the circle ' r' = 4

Step-by-step explanation:

Explanation:-

Given circle equation is x² + y² +10 x + 20 y +109 =0

x² +10 x + y² + 20 y +109 =0

x² +2 (5) (x)+(5)² -(5)²+ y² +2(10) y + (10)²-(10)² +109 =0

By using formula

(a+b)² = a² + 2 a b + b²

(x + 5 )² + ( y + 10 )² - 25 - 100 + 109 = 0

(x + 5 )² + ( y + 10 )² - 16 = 0

(x + 5 )² + ( y + 10 )² = 16

(x + 5 )² + ( y + 10 )² = (4)²

The standard equation of the circle ( x - h )² + ( y -k)² = r²

Center of the circle ( h, k) = ( -5 , -10)

Radius of the circle ' r' = 4

Conclusion:-

The equation of the circle (x + 5 )² + ( y + 10 )² = (4)²

Center of the circle ( h, k) = ( -5 , -10)

Radius of the circle ' r' = 4

3 0

3 years ago

Simplify: cos2x-cos4 all over sin2x + sin 4x

GrogVix [38]

Answer:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

$\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$ and

$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

We get:

$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}$

$=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}$

$=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}$

$=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

$=\tan\left(x\right)$

Hence final answer is

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

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3 years ago

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aniked [119]

Answer:

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