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ad-work [718]
3 years ago
13

Worksheet 1. −3|2x + 6| = −12 2. |x − 2| + 10 = 12

Mathematics
1 answer:
Misha Larkins [42]3 years ago
4 0
Answer:

1.

-3|2x+6|=-12
|2x+6|=4

2x+6=4

2x+6=-4

x1=-5, x2=-1

2.

|x-2|+10=12
|x-2|=12-10
|x-2|=2

x-2=2

x-2=-2

x1=0, x2=4
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According to the properties of inequality, when should the inequality symbol be reversed when solving a two-step any quality in
Marina86 [1]

According to the properties of inequality, we reverse the inequality symbol when we add or multiply by any negative number

Given inequality is Ax + B > C

In the first step, we subtract B from both sides

Ax + B > C

Ax + B - B > C - B

Ax > C - B

Now to remove A we divide by A on both sides

If A is negative then the reverse the inequality sign

When A is negative then > symbol becomes < symbol

so the inequality symbol be reversed when the value of A is negative

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3 years ago
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Suppose that each semester at a particular community college Manuel has to pay $1478 in tuition and $121 in fees.
Kamila [148]

Answer:

Step-by-step explanation:

<em>Let x be time, in semesters, remaining before Manuel graduates</em>.  We find that Manuel's expenses are:

$1478/semester in tuition, and

$121/semester in fees

1.  <u>x semesters remaining:  fees</u>

  Remaining cost for fees, y, is the product of the fee times the number of semesters, x:

y =($121/semester)*x

y = ($121)*x

2.  <u>x semesters remaining:  tuition</u>

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3.  <u>x semesters remaining:  tuition and fees</u>

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1 year ago
Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck
Nimfa-mama [501]
\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}
\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}

Now, isolate the variables, so you can integrate.
(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}
\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}
2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C


4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C
y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C


y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}
y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}

At x = 1, y = 0.
0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}
-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}

4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0
\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}


4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C
0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C
C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}
C = -\frac{\pi}{2}

\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}
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3 years ago
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sertanlavr [38]

Answer:

<u>Challenge #1:</u>

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2x+1=3x+2

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<u>Different examples:</u>

2x+0=4x+1

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