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marin [14]
3 years ago
13

ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. prove that ∆ ABC ≅ ∆DEF

Mathematics
1 answer:
Olegator [25]3 years ago
5 0

Answer:

Triangles ABC and DEF have the following characteristics :B and E are right angles A=D, BC=EF which congruence theorem can be used to prove ABC =DEF

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Which of the following best describes using a slope of -5/10 on a graph?
valina [46]

Answer:

move down five and move right ten

7 0
3 years ago
Easy 10 points and brainliest :)
Leya [2.2K]

Answer:

Quadrant IV

Step-by-step explanation:

If x > 0 the we are in quadrants 1 or 4

If Y < 0 we are in quadrants 3 or 4

Since both are true, we are in quadrant 4

6 0
4 years ago
Read 2 more answers
Solve the system using substitution.
Maru [420]

Answer:

Given: The following system:

4x + y − 2z=18          ......[1]

2x-3y+3z = 21          ......[2]

x-3y=6                      ......[3]

we can write equation [3] as;

3y = x-6                     ......[4]

Multiply by 3 in equation [1] to both sides of an equation we get,

3 \cdot(4x+y-2z)= 3\cdot 18 or

12x+3y-6z=54              ......[5]

Substituting the equation [4] in [2] and [5] we get;

2x-(x-6)+3z=21 or

2x-x+6+3z=21

Simplify:

x+3z=15            .....[6]                       [combine like terms]

12x+x-6-6z =54

Simplify:

13x-6z=60         ......[7]        [Combine like terms]

On Solving equation [6] and [7] simultaneously,

x+3z=15

13x-6z=60

we get the value of x

i.e, x=6

Substitute the value of x in equation x+3z=15  we get

6+3z=15 or

3z=9

Simplify:

z=3

Also, substitute the value of x=6 in equation [3] we get the value of y;

x-3y=6

6-3y=6 or

-3y = 0

Simplify:

y = 0

Therefore, the solution to the system of three linear equation is, (6, 0 , 3)


5 0
4 years ago
How many different ways can the letters of MOM be arranged
musickatia [10]
I believe 3. MOM, MMO, and OMM
5 0
3 years ago
\[f(x)=\sqrt{x ^4-16x ^{2}}\]
marusya05 [52]
f ( x ) =  \sqrt{ x^{4} -16 x^{2} }
a ) The domain:
x^4 - 16 x² ≥ 0
x² ( x² - 16 ) ≥ 0
x² - 16 ≥ 0
x² ≥ 16
x ∈ ( - ∞, - 4 ] ∪ [ 4 , + ∞ )
b ) f ` ( x ) = \frac{1}{2 \sqrt{x ^{4}-16 x^{2}  } } *  ( x^{4}-16 x^{2} )` =
= ( 2 x³ - 16 x ) / √(x^4 - 16 x²)
c ) The slope of the tangent line at x = 5:
f ` ( 5 ) = ( 2 * 125 - 16 * 5 ) / √ ( 625 - 400 ) = 170 / 15 = 34 / 3
The slope of the line normal to the graph at x = 5:
m = - 3 / 34 
4 0
3 years ago
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