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9966 [12]
3 years ago
8

Which congruence theorem can be used to prove △ABC ≅ △DBC?

Mathematics
1 answer:
Tema [17]3 years ago
7 0

Answer:

SAS

Step-by-step explanation:

Hope this helps :)

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A circle whose area is 4 has a radius of x find the area of a circle whose radius is 3x
____ [38]
The first one is 1.12867 because you need to set it up like this.
4=3.14(x)squared
4/3.14=x(squared)
x=1.12867

The second one is done simply by multiplying it out.
A=3.14(3x)squared
A=3.14(9x[squared])
A=28.26x[squared]        -The squared is referring to the x.
7 0
3 years ago
Read 2 more answers
What are the coordinates of the x-intercept(s) of the graph of y=(x-6)(x+5)
MissTica

Answer:

(-5 , 0) and (6 , 0)

Step-by-step explanation:

In order to get the coordinates of the x-intercept(s) of the graph of y=(x-6)(x+5)

we need to solve for x the equation (x-6)(x+5) = 0

(x-6)(x+5) = 0

⇌

x - 6 = 0  or  x + 5 = 0

⇌

x = 6  or  x = -5

therefore

the coordinates are :

(-5 , 0) and (6 , 0)

__________________________

:)

3 0
3 years ago
Whats 87 + 72 tenths
Alchen [17]

Answer:

94.2

Step-by-step explanation:

google and simple addition

8 0
3 years ago
Read 2 more answers
PLEASE PLEASE HELP ME
Galina-37 [17]

Answer:

  • 1. 110°

Step-by-step explanation:

Given angle - 610°

Coterminal angle is an angle in the interval 0°- 360° added or subtracted to multiples of 360°:

  • -610° + 2*360° = -610° + 720° = 110°

Correct option is 1.

7 0
3 years ago
Pls can someone help me<br><img src="https://tex.z-dn.net/?f=%20log10%20%5Csqrt%7B36%7D%20%20%2B%20%20%20%20log10%20%20%5Csqrt%7
Firlakuza [10]

Answer:

= log_{10}(\frac{6\sqrt{2} }{\sqrt{7} } )

Step-by-step explanation:

Given the expression log10\sqrt{36} + log10\sqrt{2}  - log10\sqrt{7} \\

According to the law of logarithm;

loga + log b = log(ab) and log a - log b = log(a/b)

The given expression becomes;

log10\sqrt{36} + log10\sqrt{2}  - log10\sqrt{7} \\= log_{10}(\frac{\sqrt{36} \times \sqrt{2} }{\sqrt{7} })\\= log_{10}(\frac{6\sqrt{2} }{\sqrt{7} } )

6 0
3 years ago
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