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kupik [55]
3 years ago
8

Susan needs to simplify the expression shown below.

Mathematics
1 answer:
mash [69]3 years ago
3 0

Answer:

The answer to 75 - ( 8 + 45 ÷ 3 ) × 2= 29

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What is the product of yx^2 and (6y^2x-yx)
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6y^3x^3-y^2x^3

Step-by-step explanation:

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Is 0.6 greater or less than 0.60
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It is the same. Even though you put a 0 in the end can't change the number
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1) A line with gradient 2 passes through (0, 4). Write down the equation of the line.
blondinia [14]

<em>For</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>image</em><em>,</em><em> </em><em>note</em><em> </em><em>that</em><em> </em><em>slope</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>thing</em><em> </em><em>as</em><em> </em><em>gradient</em><em>.</em>

1) y = 2x + 4

2) Substituting in x = 1 and y = 1,

1 = 4(1) + c

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So, the equation is y = 4x - 3

7 0
2 years ago
Convert the rectangular coordinates (-9, 3V3) into polar form. Express the angle
Whitepunk [10]

Answer:

(6\sqrt{3},\,\frac{5\pi}{6})

Step-by-step explanation:

The radius  r  can be found from the relationship

 r^2=x^2+y^2\\r^2=(-9)^2+(3\sqrt{3})^2\\r^2=81+27=108\\r=\sqrt{108}\\r=6\sqrt{3}

The point is in Quadrant II (-, +), so use the inverse cosine function to find the angle.

\cos{\theta}=\frac{x}{r}=\frac{-9}{6\sqrt{3}}\\\cos{\theta}=-\frac{9}{6\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\\cos{\theta}=-\frac{9\sqrt{3}}{6\cdot3}\\\cos{\theta}=-\frac{\sqrt{3}}{2}\\\\\cos^{-1}\frac{-\sqrt{3}}{2}}=\frac{5\pi}{6}

See the attached image.

7 0
3 years ago
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