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Andreas93 [3]
2 years ago
12

, 125^{5x-3} = 25^{-5x+2}" alt="7^{x} , 125^{5x-3} = 25^{-5x+2}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Vinvika [58]2 years ago
8 0
The answer is 7 13/25
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By SAA conjecture, determine which triangles are congruent.
s344n2d4d5 [400]

Answer:

The correct option is;

[B] ΔABF ≅ ΔEDF

Step-by-step explanation:

GIven that ∠FAE is congruent to ∠FEA

Therefore, triangle ΔFAE is an isosceles triangle (Definition of isosceles triangle)

Segment FA is equal to segment FE (Equal segments of isosceles triangle)

Angle ∠EFD is congruent to angle ∠AFB (Vertically opposite angles)

Therefore, triangle ΔABF is congruent to triangle ΔEDF (Angle Angle Side AAS rule of congruency)

The correct option is ΔABF ≅ ΔEDF.

6 0
3 years ago
A circle has a radius of 7x 9 y 5 cm. The area of a circle can be found using A = πr 2 . What is the area of this circle in squa
Natalka [10]

Answer:

hi

Step-by-step explanation:

Area of a circle  

A

=

π

r

2

Given  

r

=

6

x

9

y

5

c

m

A

=

π

(

6

x

9

y

5

)

2

=

36

π

x

18

y

10

s

q

c

m

A

=

113.0973

x

18

y

10

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2

7 0
3 years ago
The record low temp in Fargo, ND is -37*F. The record high is 109*F. What is the difference in the record high and the record lo
yuradex [85]

Answer:

109 - -37 = 146

Step-by-step explanation:

since you are subtracting a negative, the negative number and the sign cancel each other out. So all you have to do is add.

6 0
2 years ago
Please help me i really need help please
lesya692 [45]

Answer:A?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The radius of one circle is three the time radius of another circle. The sum of their areas is 12pi. Find the radius of each cir
zlopas [31]
   
\displaystyle\\
\begin{cases}
R_1 = 3R_2\\
\pi R_1^2 +\pi R_2^2=12\pi ~~~\Big|~~:\pi 
\end{cases} \\  \\ 
\begin{cases} 
R_1 = 3R_2\\
R_1^2 + R_2^2=12 
\end{cases} \\  \\ 
(3R_2)^2+ R_2^2=12 \\  \\ 
9R_2^2 + R_2^2=12 \\  \\ 
10R_2^2=12\\\\
R_2^2 = \frac{12}{10} = \frac{6}{5}  \\  \\ 
R_2 =  \boxed{\sqrt{\frac{6}{5}} } \\  \\ 
R_1 = \boxed{3\sqrt{\frac{6}{5}} }



7 0
3 years ago
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