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baherus [9]
3 years ago
10

Translate the following words into an ALGEBRAIC EXPRESSION:

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

3-2x+1

Step-by-step explanation:

If we indicate x the generic number we can write this expression

3-2x+1

when a sign of an operation is not expressed it means that the operation is multiplication

2x means 2*x  

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Refer to your math writing journal file named "Lines and Angles." What real-world examples did you describe for the three ways i
Sav [38]

Answer:

  1. Railway lines are example of parallel lines
  2. The floor and the walls of a room are example of perpendicular lines
  3. Two roads crossing at a signal can be termed as example of intersecting lines

Step-by-step explanation:

The lines can be related in following three ways

  1. Lines can be parallel
  2. Lines can be perpendicular
  3. Lines can be intersecting at an angle other than 90.

Now three real life examples of above three scenarios are described below:

  1. Railway lines are example of parallel lines
  2. The floor and the walls of a room are example of perpendicular lines
  3. Two roads crossing at a signal can be termed as example of intersecting lines
6 0
3 years ago
How do you simplify that question? For algebra 1.
motikmotik
\left( \sqrt{x^{-3}} \right)^5 =  \left( \sqrt{ \frac{1}{x^3} } \right)^5= \frac{1}{ (\sqrt{x^3})^5 } = \frac{1}{ \sqrt{x^{15}} } = \frac{1}{x^7 \sqrt{x} }


<span>Explanation for the last step:</span>
\sqrt{x^{15}}= \sqrt{x^{14}*x}= \sqrt{x^{14}}* \sqrt{x} =x^{ \frac{14}{2} }* \sqrt{x} =x^7 \sqrt{x}

7 0
3 years 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}}}
6 0
3 years ago
Write the equation of a circle with center (7.0) with radius 3.
harina [27]

Answer:

2.3

Step-by-step explanation:

so the 7.0 gets dived by the 3 because the radius is half of the circle.

5 0
3 years ago
Read 2 more answers
Complete the equation of the graphed linear function in point-slope form.
vazorg [7]
The point-slope form:
y-y_1=m(x-x_1)
m - slope
x₁, y₁ - the coordinates of a point

It passes through the points (1,-2) and (2,2).
(1,-2) \\&#10;x_1=1 \\ y_1=-2 \\ \\&#10;(2,2) \\&#10;x_2=2 \\ y_2=2 \\ \\&#10;m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-2)}{2-1}=\frac{2+2}{1}=4

\boxed{y-(-2)=4(x-1)}
3 0
3 years ago
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