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3 years ago

11

The pair of linear equations 6 x + 4 y = 5 and 3 x + 2 y = 10 is *

2 answers:

Artist 52 [7]3 years ago

4 0

Answer:

i think the answer is c Consistent and dependent

Step-by-step explanation:

suter [353]3 years ago

3 0

Answer:

Inconsistent

Step-by-step explanation:

6x + 4y = 5

And

3x + 2y = 10

Are two distinct parallel lines, hence they never meet.

The system has no solution, therefore it's inconsistent

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Use the substitution x = 2 − cos θ to evaluate the integral ∫ 2 3/2 ( x − 1 3 − x )1 2 dx. Show that, for a < b, ∫ q p ( x −

MrRissso [65]

If the integral as written in my comment is accurate, then we have

$I=\displaystyle\int_{3/2}^2\sqrt{(x-1)(3-x)}\,\mathrm dx$

Expand the polynomial, then complete the square within the square root:

$(x-1)(3-x)=-x^2+4x-3=1-(x-2)^2$

$I=\displaystyle\int_{3/2}^2\sqrt{1-(x-2)^2}\,\mathrm dx$

Let $x=2-\cos\theta$ and $\mathrm dx=\sin\theta\,\mathrm d\theta$ :

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-(2-\cos\theta-2)^2}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-\cos^2\theta}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{\sin^2\theta}\sin\theta\,\mathrm d\theta$

Recall that $\sqrt{x^2}=|x|$ for all $x$ , but for all $\theta$ in the integration interval we have $\sin\theta>0$ . So $\sqrt{\sin^2\theta}=\sin\theta$ :

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sin^2\theta\,\mathrm d\theta$

Recall the double angle identity,

$\sin^2\theta=\dfrac{1-\cos(2\theta)}2$

$I=\displaystyle\frac12\int_{\pi/3}^{\pi/2}(1-\cos(2\theta))\,\mathrm d\theta$

$I=\dfrac\theta2-\dfrac{\sin(2\theta)}4\bigg|_{\pi/3}^{\pi/2}$

$I=\dfrac\pi4-\left(\dfrac\pi6-\dfrac{\sqrt3}8\right)=\boxed{\dfrac\pi{12}+\dfrac{\sqrt3}8}$

You can determine the more general result in the same way.

$I=\displaystyle\int_p^q\sqrt{(x-a)(b-x)}\,\mathrm dx$

Complete the square to get

$(x-a)(b-x)=-(x-a)(x-b)=-x^2+(a+b)x-ab=\dfrac{(a+b)^2}4-ab-\left(x-\dfrac{a+b}2\right)^2$

and let $c=\frac{(a+b)^2}4-ab$ for brevity. Note that

$c=\dfrac{(a+b)^2}4-ab=\dfrac{a^2-2ab+b^2}4=\dfrac{(a-b)^2}4$

$I=\displaystyle\int_p^q\sqrt{c-\left(x-\dfrac{a+b}2\right)^2}\,\mathrm dx$

Make the following substitution,

$x=\dfrac{a+b}2-\sqrt c\,\cos\theta$

$\mathrm dx=\sqrt c\,\sin\theta\,\mathrm d\theta$

and the integral reduces like before to

$I=\displaystyle\int_P^Q\sqrt{c-c\cos^2\theta}\,\sin\theta\,\mathrm d\theta$

where

$p=\dfrac{a+b}2-\sqrt c\,\cos P\implies P=\cos^{-1}\dfrac{\frac{a+b}2-p}{\sqrt c}$

$q=\dfrac{a+b}2-\sqrt c\,\cos Q\implies Q=\cos^{-1}\dfrac{\frac{a+b}2-q}{\sqrt c}$

$I=\displaystyle\frac{\sqrt c}2\int_P^Q(1-\cos(2\theta))\,\mathrm d\theta$

(Depending on the interval [<em>p</em>, <em>q</em>] and thus [<em>P</em>, <em>Q</em>], the square root of cosine squared may not always reduce to sine.)

Resolving the integral and replacing <em>c</em>, with

$c=\dfrac{(a-b)^2}4\implies\sqrt c=\dfrac{|a-b|}2=\dfrac{b-a}2$

because $a$ , gives

$I=\dfrac{b-a}2(\cos(2P)-\cos(2Q)-(P-Q))$

Without knowing <em>p</em> and <em>q</em> explicitly, there's not much more to say.

7 0

3 years ago

Math Please Help Make sure you explain your answer Thanks Have a blessed day!!

ivolga24 [154]

Answer:

Option B: x = 15 is a solution, but x = 9 is not is the correct answer.

Step-by-step explanation:

Given equation is:

x - 3 = 12

The solutions we have to check are;

x = 9 and x = 15

Putting x = 9 in the given equation;

9 - 3 = 12

6 ≠ 12

Putting x = 15 in the given equation;

15 - 3 = 12

12 = 12

Hence,

Option B: x = 15 is a solution, but x = 9 is not is the correct answer.

4 0

3 years ago

Liono4ka [1.6K]

Answer:

The answer is 150.72

Step-by-step explanation:

$v = \pi {r}^{2} \times h \\ = 3.14 \times 4 \times 4 \times 3 \\ = 150.72 {ft}^{3}$

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3 years ago

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The table below shows two equations: Equation 1 |4x − 3|− 5 = 4 Equation 2 |2x + 3| + 8 = 3 Which statement is true about the so

coldgirl [10]

| 4x - 3| - 5 = 4
| 4x - 3| = 9

4x - 3 = 9 -(4x - 3) = 9
4x = 12 -4x + 3 = 9
x = 3 -4x = 6
x = -6/4 or -1.5
solution to equation 1 : x = 3 and x = -1.5

|2x + 3| + 8 = 3
| 2x + 3| = - 5....no solution because an absolute value cannot equal a negative number

4 0

3 years ago

What is the area in square inches 3/8 inch 2/3 inche

Vlada [557]

Answer:

0.25 in²

Step-by-step explanation:

Repeat you last class this is so basic...

6 0

4 years ago

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