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3 years ago

I really hope someone can hekp me with my hwit's linear equations...

Mathematics

1 answer:

olchik [2.2K]3 years ago

6 0

You first should isolate the "y" in the equation.

3x - 2y = 6 Subtract 3x on both sides

3x - 3x - 2y = 6 - 3x

-2y = 6 - 3x Divide -2 on both sides to get "y" by itself

$y = \frac{6-3x}{-2}$

$y = -3 + \frac{3}{2} x$

This equation is:

y = mx + b

"m" being the slope, "b" being the y-intercept (when x = 0)

The y-intercept is -3 in the equation. So you can eliminate choices A and C.

You can plug in any of the numbers of x(-2 , -1 , 0, 1, 2) into this equation, to find it's y value.

You can plug in 1 for x:

y = -3 + $\frac{3}{2} x$

y = -3 + $\frac{3}{2} (1)$

y = -3 + $\frac{3}{2}$ Make the denominators the same

$y = \frac{-6}{2} +\frac{3}{2} = \frac{-3}{2}$

Your answer is B

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3 years ago

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borishaifa [10]

Answer:

1.) Tan C = $\frac{12}{5}$

2.) Sin Z = $\frac{24}{26}$ or $\frac{12}{13}$ if reduced

3.) Cos A = $\frac{15}{17}$

4.) Cos A = $\frac{12}{37}$

Step-by-step explanation:

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2 years ago

How is good at statics to help me? Plz

IgorC [24]

Answer:

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3 years ago

The figure below shows a rectangle ABCD having diagonals AC and DB: A rectangle ABCD is shown with diagonals AC and BD. Jimmy wr

yaroslaw [1]

Answer:

C. DC = DC. By reflexive property of equality.

Step-by-step explanation:

We are given that,

In step 4, Jimmy proved ΔADC ≅ ΔBCD by the SAS Congruence Postulate.

For that, the previous steps states,

Step 1: AD = BC, as opposite sides of a rectangle are congruent.

Step 2: ∠ADC = ∠BCD

So, in order to satisfy the SAS Postulate, we must have another pair of sides equal from the respective rectangles.

As, we can see that in ΔADC and ΔBCD, the side DC is a common side.

Thus, in the missing step, we get,

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Hence, option C is correct.

5 0

3 years ago

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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

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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3 years ago

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