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sdas [7]
3 years ago
7

HELLPPPPPP

Mathematics
2 answers:
Nady [450]3 years ago
4 0
<h3>Multiplication Property of Equality is the missing reason</h3>

<em><u>Solution:</u></em>

Given that,

five thirds times x plus 2 equals negative 3

Therefore,

\frac{5}{3}x + 2 = -3

<em><u>Multiplication Property of Equality</u></em>

The Multiplication Property of Equality means that, if both sides of an equation is multiplied by same number, the equation still remains equal

\frac{5}{3}x + 2 = -3

Multiply both sides by 3

3(\frac{5}{3}x + 2) = -3 \times 3\\\\5x + 6 = -9

<em><u>Subtraction Property of Equality</u></em>

The Subtraction Property of Equality means that, we can subtract the same number from both sides of equation and the equation will still be true

5x + 6 = -9

Subtract 6 from both sides

5x + 6 - 6 = -9  - 6

5x = -15

<em><u>Division Property of Equality</u></em>

The Division Property of Equality means that, if both sides of an equation is divided by same number, the equation still remains true

5x = -15

Divide both sides by 5

x = -3

joja [24]3 years ago
4 0

Answer: Multiplication Property Of Equality

Step-by-step explanation:

look at the image below

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Answer:

x < 9

Step-by-step explanation:

Given inequality:

6x + 9 < 63

Subtract 9 from both sides:

⇒ 6x + 9 - 9 < 63 - 9

⇒ 6x < 54

Divide both sides by 6:

⇒ 6x ÷ 6 < 54 ÷ 6

⇒ x < 9

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Write the expression below. Divide 9 by the sum of 6 and 4
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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur
Butoxors [25]
\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx

Setting x=5\sin\theta, you have \mathrm dx=5\cos\theta\,\mathrm d\theta. Then the integral becomes

\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta
\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

Now, \sqrt{x^2}=|x| in general. But since we want our substitution x=5\sin\theta to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means \theta=\sin^{-1}\dfrac x5, which implies that \left|\dfrac x5\right|\le1, or equivalently that |\theta|\le\dfrac\pi2. Over this domain, \cos\theta\ge0, so \sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta.

Long story short, this allows us to go from

\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

to

\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta

Then integrate term-by-term to get

\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)
=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C

Now undo the substitution to get the antiderivative back in terms of x.

=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C
4 0
2 years ago
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