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4 years ago

How much larger is 8 in 8237 and 8 in 823.7

2 answers:

8 0

Since the 8 in the first number is in the thousands place, its value it 8*1000=8000.
Since the 8 in the second number is in the hundreds place, its value is 8*100=800.

8000 is ten times 800.
Final answer: 10x as large

oee [108]4 years ago

6 0

From right to left the place ment system is ones tens hundreds thousands.

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The matrix below represents a system of equations.

UkoKoshka [18]

Answer:A

Step-by-step explanation:

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7 0

2 years ago

Cost of a CD: $17.50 Markup: 45%

Marizza181 [45]

Answer:

25.375

Step-by-step explanation:

17.50×45%=7.875

17.50+7.875=25.375 new price

8 0

2 years ago

Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

andreev551 [17]

Recall the angle sum identity for cosine:

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

==> sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))

Then rewrite the equation as

sin(4x) sin(5x) + sin(4x) sin(3x) - sin(2x) sin(x) = 0

1/2 (cos(-x) - cos(9x)) + 1/2 (cos(x) - cos(7x)) - 1/2 (cos(x) - cos(3x)) = 0

1/2 (cos(9x) - cos(x)) + 1/2 (cos(7x) - cos(3x)) = 0

sin(5x) sin(-4x) + sin(5x) sin(-2x) = 0

-sin(5x) (sin(4x) + sin(2x)) = 0

sin(5x) (sin(4x) + sin(2x)) = 0

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

Rewrite the equation again as

sin(5x) (2 sin(2x) cos(2x) + sin(2x)) = 0

sin(5x) sin(2x) (2 cos(2x) + 1) = 0

sin(5x) = 0 or sin(2x) = 0 or 2 cos(2x) + 1 = 0

sin(5x) = 0 or sin(2x) = 0 or cos(2x) = -1/2

sin(5x) = 0 ==> 5x = arcsin(0) + 2nπ or 5x = arcsin(0) + π + 2nπ

… … … … … ==> 5x = 2nπ or 5x = (2n + 1)π

… … … … … ==> x = 2nπ/5 or x = (2n + 1)π/5

sin(2x) = 0 ==> 2x = arcsin(0) + 2nπ or 2x = arcsin(0) + π + 2nπ

… … … … … ==> 2x = 2nπ or 2x = (2n + 1)π

… … … … … ==> x = nπ or x = (2n + 1)π/2

cos(2x) = -1/2 ==> 2x = arccos(-1/2) + 2nπ or 2x = -arccos(-1/2) + 2nπ

… … … … … … ==> 2x = 2π/3 + 2nπ or 2x = -2π/3 + 2nπ

… … … … … … ==> x = π/3 + nπ or x = -π/3 + nπ

(where n is any integer)

5 0

3 years ago

Please please help me and fast

Eduardwww [97]

Its 12. Part a was 2 part b was 3
6×2=12
4×3=12

4 0

3 years ago

Sabendo que "K" satisfaz a equação {2(k-8) + 3(-k+1) = -4k +11} Os valores reais de x que satisfazem a equação 15x² - kx + 1 = 0

erastova [34]

Answer:

D) 1/5 e 1/3

Step-by-step explanation:

You have the following quadratic equation:

$15x^2-kx+1=0$ (1)

In order to find the values of x that are solution to the equation (1), you first find the solution for k in the following equation:

$2(k-8)+3(-k+1)=-4k+11\\\\2k-16-3k+3=-4k+11\\\\2k-3k+4k=11+16-3\\\\3k=24\\\\k=8$

Next, you replace the previous value of k in the equation (1) and you use the quadratic formula to find the roots:

$15x^2-8x+1=0\\\\x_{1,2}=\frac{-(-8)\pm \sqrt{(-8)^2-4(15)(1)}}{2(15)}\\\\x_{1,2}=\frac{8\pm 2}{30}\\\\x_1=\frac{1}{5}\\\\x_2=\frac{1}{3}$

Then, the roots of the equation (1) are

D) 1/5 e 1/3

5 0

4 years ago