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saw5 [17]
3 years ago
5

What is the name of the point where the three altitudes of a triangle intersect?

Mathematics
2 answers:
san4es73 [151]3 years ago
6 0

Answer:

D point of conncurrency of the altitudes

the second one i am not sure of but that is the first part to your answer

Step-by-step explanation:

kenny6666 [7]3 years ago
6 0

1. D. orthocenter is the answer.

2. C. 30 degrees

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If y = 3x - 4 and the domain is {-3,-1,4}, find the range.
GREYUIT [131]
We could put the domain into the x values to get the y values (range). 

y = 3(-3) - 4
y = -9 - 4
y = -13

y = 3(-1) - 4
y = -3 - 4
y = -7

y = 3(4) - 4
y = 12 - 4
y = 8
4 0
3 years ago
Read 2 more answers
Need help finding this answer
Cloud [144]

Answer: The first option.


Step-by-step explanation:

1. The formula for calculate the perimeter is:

P=2l+2w

Where l is the lenght and w is the width.

2. The first perimeter is:

P_1=2(105m)+2(12.5m)=235m

3. If you multiply the dimension by a scale of factor of 2, is the same as write the following expression:

P_2=2(2l+2w)

Where 2l+2w is the first perimeter.

4. Then:

 P_2=2(235m)=470m



4 0
3 years ago
Read 2 more answers
Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \triangle ADC \cong \triangle BDC (LL)

5) \overline{AC} \cong \overline{BC} (CPCTC)

<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

8 0
1 year ago
What doexs this equal: 3ab-9ab+7ab
AfilCa [17]

Answer:

I think that it equals 1ab. Because 3ab - 9ab= -6ab + 7ab= 1ab



8 0
3 years ago
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Seth measured a hotel and made a scale drawing the the scale he used was 1 inch : 3 feet the actual of the hotel is 18 feet how
sertanlavr [38]

Answer:

Step-by-step explanation:

Width of the room in the drawing = x

1 : 3 :: x : 18

Product of means = Product of extremes

         3 * x =  18 *1

             x = 18 ÷ 3

            x = 6

Width of the room in the drawing = 6 inches

7 0
2 years ago
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