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yarga [219]
3 years ago
13

Divide: 2 ÷ 1/2A. 2B. 16C. 8D. 4​

Mathematics
1 answer:
laiz [17]3 years ago
4 0

Answer:

D

Step-by-step explanation:

hope this helps :)

if this did help, mark me brainliest

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Given: ΔABC, where AB = BC Prove: m∠BAC = m∠BCA Statement Reason 1. Let ΔABC be an isosceles triangle with AB = BC. given 2. Cre
weqwewe [10]

Answer:

D. SAS

Step-by-step explanation:

Given: ΔABC

Bisecting <ABC to create point D implies that BD is a common side to  ΔABD and ΔCBD.

Also,

m<ABD = m<CBD (angle bisector)

BA = BC (given property of the isosceles triangle)

Therefore,

ΔABD ≅ ΔCBD (Side Angle Side)

The reason for statement 5 in this proof is that ΔABD ≅ ΔCBD by SAS (Side-Angle-Side) relations of the congruent triangles.

4 0
3 years ago
......................................
pantera1 [17]
your answer is 41.38
4 0
3 years ago
a small box in the shape of a cube has a length of 5 inches. how many cubic inches will the box hold?
Mama L [17]

Answer:

125 cubic inches

Step-by-step explanation:

volume of cube is l^3 so 5^3 is 125.

6 0
2 years ago
Read 2 more answers
This pattern repeats every 4 terms, and the rule is triangle, plus, star, circle. What is the shape in the 16th term of this pat
Alexxandr [17]

Answer:

Fill in the blanks to make the following statements true. Also named the property used.  

a. (-5) + (-4) = ___ + (-5),

b. 4 + __ = 4,

c. - 53 + ___ = - 53,  

d. 4 + [(-5) + (7)] = [4 + (7)] + ___,

e. 25 + [(-50) + 5 ] = (25 + 5) + ___,

f. (-4) + ___ = -4,

g. 4 + (-4) = ___,

h. 5 + ___ = 0 ,

Step-by-step explanation:

Fill in the blanks to make the following statements true. Also named the property used.  

a. (-5) + (-4) = ___ + (-5),

b. 4 + __ = 4,

c. - 53 + ___ = - 53,  

d. 4 + [(-5) + (7)] = [4 + (7)] + ___,

e. 25 + [(-50) + 5 ] = (25 + 5) + ___,

f. (-4) + ___ = -4,

g. 4 + (-4) = ___,

h. 5 + ___ = 0 ,

5 0
3 years ago
Prove that is here<br><img src="https://tex.z-dn.net/?f=1%20-%20cos%20%7B2%7Da%20%5Cdiv%201%20-%20sin%20a%7B2%7D%20%20%3D%20tan%
garri49 [273]

\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}

<h3>LHS</h3>

\boxed{\sf \dfrac{cosA}{sinA}=cotA}

\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}

\\ \sf\longmapsto 1-cot2A

\\ \sf\longmapsto 1-\dfrac{1}{tan2A}

\\ \sf\longmapsto \dfrac{tan2A-1}{tan2A}

\\ \sf\longmapsto tan2A

8 0
3 years ago
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