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LenKa [72]
3 years ago
12

What is the image point of (-6,-9)(−6,−9) after a translation left 1 unit and down 5 units?

Mathematics
1 answer:
Anna [14]3 years ago
7 0

Answer:

(-7,-4)

Step-by-step explanation:

-6-1=-7

-9-(-5)=-4

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*BRAINLIEST* (Easy Question: Warm UP) Find a positive solution to each equation: x2 = 9/4?
prisoha [69]

Answer:

<u>A</u>

Step-by-step explanation:

<u>Given</u>

  • x² = 9/4

<u>Solving</u>

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  • ⇒ x = 3/2
  • <u>Option A is the correct option.</u>
7 0
2 years ago
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Find f.<br><br> A) 7.4<br><br> B) 8.2<br><br> C) 10.5<br><br> D) 11.1
tatyana61 [14]
F=72

g=6

------------

\cos { \left( F \right)  } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }

Therefore:

\cos { \left( 72 \right)  } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right)  } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }

\\ \\ 12e\cdot \cos { \left( 72 \right)  } ={ e }^{ 2 }+36-{ f }^{ 2 }\\ \\ \therefore \quad { f }^{ 2 }={ e }^{ 2 }-12e\cdot \cos { \left( 72 \right)  } +36\\ \\ \therefore \quad f=\sqrt { { e }^{ 2 }-12e\cdot \cos { \left( 72 \right) +36 }  } \\ \\ \therefore \quad f=\sqrt { e\left( e-12\cos { \left( 72 \right)  }  \right) +36 }

But what is e?

E=76

G=32

g=6

And:

\frac { e }{ \sin { \left( E \right)  }  } =\frac { g }{ \sin { \left( G \right)  }  }

Which means that:

\frac { e }{ \sin { \left( 76 \right)  }  } =\frac { 6 }{ \sin { \left( 32 \right)  }  } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  }

If you take this value into account, you will discover that f is...

f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  } \left( \frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  } -12\cos { \left( 72 \right)  }  \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)

So I would have to say that the answer is approximately (c).
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3 years ago
Round 0.172 to the nearest hundredth.
aleksandr82 [10.1K]
It is <span>Round 0.172 to the nearest hundredth.</span>
8 0
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If two tangent segments to a circle share a common endpoint outside a circle, then the two segments are
matrenka [14]

If two tangent segments to a circle share a common endpoint outside a circle, then the two segments are congruent. This is according to the intersection of two tangent theorem. The theorem states that  given a circle, if X is any point within outside the circle and if Y and Z are points such that XY and XZ are tangents to the circle, then XY is equal to XZ.

<span> </span>

4 0
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Which pair of triangles can be proven congruent by the HL theorem?<br><br><br> a, b, or c?
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Answer:

C

Step-by-step explanation:

The HL Theorem states; If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent

6 0
2 years ago
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