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4 years ago

The value of the 7 in 27,459 is how many times the value of the 7 in 40,735

Mathematics

2 answers:

Ksivusya [100]4 years ago

7 0

Ten times greater you're welcome

garri49 [273]4 years ago

6 0

It is ten times greater

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Cos^2 x+4sin^2 x/2=1

lana [24]

$Let\ \dfrac{x}{2}=a,\ therefore\ x=2a.\\\\\cos^2x+4\sin^2\dfrac{x}{2}=\cos^22a+4\sin^2a\\\\\text{use}\ \cos2x=\sin^2x-\cos^2x\\\\=(\sin^2a-\cos^2a)^2+4\sin^2a\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(\sin^2a)^2-2(\sin^2a)(\cos^2a)+(\cos^2a)^2+4\sin^2a\\\\=\sin^4a-2\sin^2a\cos^2a+\cos^4a+4\sin^2a\\\\=\underbrace{\sin^4a+2\sin^2a\cos^2a+\cos^4a}_{(*)}-4\sin^2a\cos^2a+4\sin^2a\\\\\text{use}\ (*)\qquad(a+b)^2=a^2+2ab+b^2$

$=\underbrace{(\sin^2a)^2+2\sin^2a\cos^2a+(\cos^2a)^2}_{(*)}-4\sin^2a(\cos^2a-1)\\\\=(\sin^2a+\cos^2a)^2-4\sin^2a(\cos^2a-1)\\\\\text{use}\ \sin^2a+\cos^2a=1\to\sin^2a=\cos^2a-1\\\\=1^2-4\sin^2a(\sin^2a)=1-4\sin^4a=1-(2\sin^2a)^2$

$\cos^22a+4\sin^2a=1\\\\1-(2\sin^2a)^2=1\qquad\text{subtract 1 from both sides}\\\\-(2\sin^2a)^2=0\to2\sin^2a=0\qquad\text{divide both sides by 2}\\\\\sin^2a=0\to\sin a=0\\\\a=k\pi\ for\ k\in\mathbb{Z}\\\\\dfrac{x}{2}=k\pi\qquad\text{multiply both sides by 2}\\\\\boxed{x=2k\pi\ for\ k\in\mathbb{Z}}$

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3 years ago

determine if the following set of ordered pairs represents a quadratic function explain (5,7),(7,11),(9,14),(11,18)

Fynjy0 [20]

Answer:the pairs are not a quadratic eqation

Step-by-step explanation:

The differences between the differences of the y value are not consistent

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3 years ago

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elena55 [62]

Answer:

32x-4

Step-by-step explanation:

1. Add the numbers.

2x(16)-4

2. Multiply the numbers.

32x-4

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3 years ago

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There are 24 red roses and 36 white roses in the bouquet. what is the ratio of red roses to white roses

Lana71 [14]

God is always the answer...always

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3 years ago

Find the standard form polynomial that represents this product. (6 - y - 4y^2)(-5 + 7y^2)

Tema [17]

Answer:

-28y^4 - 7y^3 + 62y^2 + 5y - 30

Step-by-step explanation:

Opening the parenthesis and simplifying by combining like terms:

(6 - y - 4y^2)(-5 + 7y^2) =
6(-5) +6(7y^2) - y(-5) -y(7y^2) -4y^2(-5) -4y^2(7y^2) =
-30 + 42y^2 + 5y - 7y^3 +20y^2 -28y^4 =
-28y^4 - 7y^3 + 62y^2 + 5y - 30

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4 years ago