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laiz [17]
3 years ago
6

ZABD and ZDBC are supplementary angles.

Mathematics
2 answers:
Dominik [7]3 years ago
7 0

Answer:

Solution given:

x+120°=180°[linear pair][supplementary]

x=180°-120=60°

x=60° is your answer

Mila [183]3 years ago
4 0
<h3>Answer:</h3>

<u>As above angles are supplementary</u>,

  • x + 120° = 180°
  • x = 180° - 120°
  • x = 60°

<u>Hence, value of x is 60°</u>.

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vovangra [49]

Answer:

1. 27^{\frac{2}{3} } =9

2. \sqrt{36^{3} } =216

3. (-243)^{\frac{3}{5} } =-27

4. 40^{\frac{2}{3}}=4\sqrt[3]{25} =4325

5. Step 4: (\frac{343}{27}) ^{-1} =\frac{27}{343}

6. D. -72cd^{7}

Step-by-step explanation:

Use the following properties:

a^{\frac{x}{y} } =\sqrt[x]{a^{y} }

\sqrt[n]{ab} =\sqrt[n]{a} \sqrt[n]{b}

a^{-n} =\frac{1}{a^{n} }

(xy)^{z} =x^{z} y^{z} \\\\

(x^{y}) ^{z} =x^{yz}

x^{y} x^{z} =x^{y+z}

So:

1. 27^{\frac{2}{3} } =\sqrt[3]{27^{2}} =\sqrt[3]{729} }=9

2. \sqrt{36^{3} } =\sqrt{36*36*36} =\sqrt{36} \sqrt{36}  \sqrt{36} =6*6*6=216

3. (-243)^{\frac{3}{5} } =\sqrt[5]{-243^{3} } =\sqrt[5]{-14348907} =-27

4. 40^{\frac{2}{3}}=\sqrt[3]{40^{2} } =\sqrt[3]{2^{6} 5^{2} } =\sqrt[3]{2^{6} } \sqrt[3]{5^{2} } =2^{\frac{6}{3} } 5^{\frac{2}{3} } =4 *5^{\frac{2}{3} } =4\sqrt[3]{5^{2} } =4\sqrt[3]{25}=4325

5. (\frac{343}{27}) ^{-1} =\frac{1}{\frac{343}{27} } =\frac{27}{343}

6.

(-8c^{9} d^{-3} )^{\frac{1}{3} } *(6c^{-1}d^{4})^{2} =\sqrt[3]{-8} c^{3} d^{-1} 36c^{-2} d^{8} \\\\-2c^{3} d^{-1} 36c^{-2} d^{8}=-72cd^{7}

7 0
3 years ago
A triangle has one angel that measures 51 degrees and one angle that measures 39 degrees. what kind of triangle is it?
Lostsunrise [7]
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3 0
3 years ago
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A set of golf clubs is on sale for $150.00, which is 70% off its original price.What was the original price of the golf clubs
irina1246 [14]

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3 years ago
Find the equation of a line that is perpendicular to line g that contains (P, Q).
Nat2105 [25]

We want to find the equation of a line that is perpendicular to G and pass through point (P, Q)

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Point 2 = (x2,y2) = (-3,2)

So, let find the slope of this line G.

Slope can be calculated using

m = ∆y/∆x

m = (y2 - y1) / (x2 - x1)

m = (2-6) / (-3--2), -×- = +

m = -4 / (-3 + 2)

m = -4 / -1

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So, the gradient or slope of line G is 4

From geometry, since the equation of the line we want to find is perpendicular to the G,

Then, the slope of the line is

m2 = -1 / m1

m2 = - 1 / 4

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Then, also we can write equation of a line given the slope and a point using

m = (y - y1) / (x - x1)

Where m = -¼

And the point (x1,y1) = (P, Q)

Then, we have

-¼ = (y - Q) / (x - P)

Cross multiply

-(x - P) = 4(y - Q)

-x + P = 4y - 4Q

Rearrange

-x - 4y = - 4Q - P

Divide through by -1

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Ainat [17]

Answer:

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