Answer:
Solution given;
<ABD=<BAC+<ACB
<u>Since</u><u> </u><u>exterior</u><u> </u><u>angle</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>triangle</u><u> </u><u>is</u><u> </u><u>equal</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>sum</u><u> </u><u>of</u><u> </u><u>two</u><u> </u><u>opposite</u><u> </u><u>interior</u><u> </u><u>angle</u>
26x+20=19x-15+9x+25
solve like terms
26x+20=28x+10
subtracting both by 10
26x+20-10=28x+10-10
Subtracting both side by 26x
10=28x-26x
2x=10
dividing both side by 2
2x/2=10/2
x=5
Now
<ABD=26*5+20=l50°
<u>T</u><u>h</u><u>e</u><u> </u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u> </u><u>o</u><u>f</u><u> </u><u><</u><u>A</u><u>B</u><u>D</u><u> </u><u>i</u><u>s</u><u> </u><u>1</u><u>5</u><u>0</u><u>°</u>
Answer: 5, 3, 2, 4, 1
<u>Step-by-step explanation:</u>
The proof should be written as follows:
Statement Reason
1. ABCD is a parallelogram 1. Given
2. AB||CD and AC||BD 2. Definition of a parallelogram
3. ∠EBD ≅ ∠CAB and ∠FAC ≅ ∠ABD 3. Corresponding Angles Thm
4. ∠EBD ≅ ∠CDB and ∠ACD ≅ ∠FAC 4. Transitive Property of Cong
5. ∠CDB ≅ ∠CAB and ∠ACD ≅ ∠ABD 5. Alternate Interior Angles Thm
This is the order provided in your question:
5. ∠CDB ≅ ∠CAB and ∠ACD ≅ ∠ABD 5. Alternate Interior Angles Thm
3. ∠EBD ≅ ∠CAB and ∠FAC ≅ ∠ABD 3. Corresponding Angles Thm
2. AB||CD and AC||BD 2. Definition of a parallelogram
4. ∠EBD ≅ ∠CDB and ∠ACD ≅ ∠FAC 4. Transitive Property of Cong
1. ABCD is a parallelogram 1. Given
<em>Note: #2 & #4 might be reversed</em>
so the investigator found the skid marks were 75 feet long hmmm what speed will that be?
nope, the analysis shows that Charlie was going faster than 35 m/h.
now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been
Answer:
-19
Step-by-step explanation:
Answer:
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Work Shown:
The answer has been confirmed with WolframAlpha which is basically a calculator app, but does a ton more than any average calculator.
The CAS (computer algebra system) in the app GeoGebra is another tool I use all the time.
Yet another way to confirm the answer is to graph the original expression and the answer as separate functions. You should find they produce the <u>exact</u> same curve. One curve perfectly overlaps the other. This visually verifies the two expressions are the same thing, just written a different way of course.
Let me know if you have any questions about any step. The basic idea I did was to pull out the GCF and then combine like terms.