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jasenka [17]
3 years ago
9

Plzz help y+10 /10 = y-2 /4 solve for y

Mathematics
1 answer:
user100 [1]3 years ago
7 0

Answer:

There are no values of y in this equation. so therefore there is no solution.

Step-by-step explanation:

maybe you mad a mistake in the question please make sure this is the correct question :) thanks and have a nice day

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(I need the answer right now please) ΔA'B'C' was constructed using ΔABC and line segment EH. 2 triangles are shown. Line E H is
irina1246 [14]

Answer:

The true statements are

1. BD = DB'

3. m∠EFA = 90°

4. The line of reflection, EH, is the perpendicular bisector of BB', AA', and

   CC'

Step-by-step explanation:

* Lets explain how to solve the problem

- Reflection is flipping an object over the line of reflection.

- The object and its image have the same shape and size, but the

  figures are in opposite directions from the line of reflection

- The objects appear as if they are mirror reflections, with right and left

  reversed

- The line of reflection is a perpendicular bisector for all lines joining

  points on the figure with their corresponding images

- Look to the attached figure for more understand

* Lets solve the problem

- ΔA'B'C' was constructed using ΔABC and line segment EH, where

 EH is the reflection line

- D is the mid-point of BB'

- F is the mid-point of AA'

- G is the mid-point of CC'

* Lets find from the answer the true statements

1. BD = DB'

∵ D is the mid point of BB'

- Point D divides BB' into two equal parts

∴ BD = DB' ⇒ <em>True</em>

2. DF = FG

- It depends on the size of the sides and angles of the triangle

∵ We can't prove that

∴ DF = FG ⇒ <em>Not true</em>

3. m∠EFA = 90°

∵ The line of reflection ⊥ the lines joining the points with their

   corresponding images

∴ EH ⊥ AA' and bisect it at F

∴ m∠EFA = 90° ⇒ <em>True</em>

4. The line of reflection, EH, is the perpendicular bisector of BB',

   AA', and CC'

- Yes the line of reflection is perpendicular bisectors of them

∴ The line of reflection, EH, is the perpendicular bisector of BB',

   AA', and CC' ⇒ <em>True</em>

5. ΔABC is not congruent to ΔA'B'C'

∵ In reflection the object and its image have the same shape and size

∴ Δ ABC is congruent to Δ A'B'C'

∴ ΔABC is not congruent to ΔA'B'C' ⇒ <em>Not true</em>

4 0
3 years ago
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What is PIE multiplied by PIE?
oksian1 [2.3K]
Pie equal 3.14. So therefore 3.14 x 3.14= 9.8596
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Decompose the figure into regions that are closest to each
Korolek [52]

The image is decomposed as follows: H1 and H2. Where original graph is Hx.

<h3>Are the images (attached) valid decompositions of the original graph?</h3>

  • Yes, they are because, H1 and H1 are both sub-graphs of Hx; also
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Hence, {H1 , H2} are valid decomposition of G.

<h3>What is a Graph Decomposition?</h3>

A decomposition of a graph Hx is a set of edge-disjoints sub graphs of H, H1, H2, ......Hn, such that UHi = Hx

See the attached for the Image Hx - Pre decomposed and the image after the graph decomposition.

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Use the Venn diagram to calculate probabilities. Circles A and B overlap. Circle A contains 15, circle B contains 10, and the in
grin007 [14]

Answer:

(B) P(B)=\dfrac{16}{31}

Step-by-step explanation:

From the Venn diagram, n(AUB)=31

The given probabilities in the option are calculated below:

P(A)=\dfrac{21}{31}\\\\P(B)=\dfrac{16}{31}\\\\P(A|B)=\dfrac{P(A\cap B)}{P(B)}= \dfrac{6/31}{16/31} =\dfrac{6}{16}=\dfrac{3}{8} \\\\P(B|A)=\dfrac{P(B\cap A)}{P(A)}= \dfrac{6/31}{21/31} =\dfrac{6}{21}=\dfrac{3}{7}

The only correct option is the probability of B which is \frac{16}{31}

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