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ASHA 777 [7]
4 years ago
6

Geometry Proof - Thanks in advance for the help!

Mathematics
1 answer:
RideAnS [48]4 years ago
8 0

Answer:

Using reflexive property (for side), and the transversals of the parallel lines, we can prove the two triangles are congruent.

Step-by-step explanation:

  • Since AB and DC are parallel and AC is intersecting in the middle, you can make out two pairs of alternate interior angles<em>.</em> These angle pairs are congruent because of the alternate interior angles theorem. The two pairs of congruent angles are: ∠DAC ≅ ∠BCA, and ∠BAC ≅ ∠DCA.
  • With the reflexive property, we know side AC ≅ AC.
  • Using Angle-Side-Angle theorem, we can prove ΔABC ≅ ΔCDA.
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A line has slope −34 and y–intercept 5. Which answer is the equation of the line? y=−5x+34 y=−34x+5 y=34x−5 y=5x−34?
Dafna1 [17]

The answer would be the first equation!!!!

3 0
4 years ago
Read 2 more answers
2/3, 2/8, 2/6 how to do least to greatest?
Morgarella [4.7K]

Answer: 2/8, 2/6, 2/3

Step-by-step explanation:

First, find the LCM of 3, 6, and 8, in this case 24.  Then, make the denominator of each fraction 24: 16/24, 6/24, 8/24.  Then, simply sort the fractions from least to greatest by their numerators: 6/24, 8/24, 16/24

Hope it helps <3

5 0
3 years ago
PLEASE HELP :(((((((((
ziro4ka [17]
The first box would be 3.14 then second box would be the pi symbol and the third box would be 22/7
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3 years ago
(15 pts) 4. Find the solution of the following initial value problem: y"-10y'+25y = 0 with y(0) = 3 and y'(0) = 13
jolli1 [7]

Answer:

y(x)=3e^{5x}-2xe^{5x}

Step-by-step explanation:

The given differential equation is y''-10y'+25y=0

The characteristics equation is given by

r^2-10r+25=0

Finding the values of r

r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5

We got a repeated roots. Hence, the solution of the differential equation is given by

y(x)=c_1e^{5x}+c_2xe^{5x}...(i)

On differentiating, we get

y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)

Apply the initial condition y (0)= 3 in equation (i)

3=c_1e^{0}+0\\\\c_1=3

Now, apply the initial condition y' (0)= 13 in equation (ii)

13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2

Therefore, the solution of the differential equation is

y(x)=3e^{5x}-2xe^{5x}

5 0
3 years ago
15points+brainliest!! please help i really need this done
aev [14]

Answer:

2 units right, 5 units down

8 0
3 years ago
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