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bekas [8.4K]
3 years ago
9

Two angles inside the triangle that do no share a vertex with the exterior.

Mathematics
1 answer:
Dvinal [7]3 years ago
3 0
Exterior angle I think
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If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of beta will be :​
Tems11 [23]
<h3>Given -:</h3>

change in emitter current, ∆l(E) = 4 mA

change in collector current,∆I(C) = 3.5 mA

<h3>To find -:</h3>

value of B

<h3>Solution :-</h3>

∆I(E) = ∆I(C) + ∆l(B), here ∆l(B) is the change in base current.

4 = 3.5 + ∆l(B)

∆I(B) = 0.5 mA

<h3>Now, B = ∆l (C) ÷ ∆l (B)</h3>

B = 3.5 ÷ 0.5

B = 7.0

<h3>so, the value of A = 7.0</h3>

8 0
2 years ago
A rectangle’s perimeter and its area have the same numerical value. The width of the
Zolol [24]

Step-by-step explanation:

6 units.

(3*2)+2x=3x

Set the perimeter equal to the area

x(length)=6 units

7 0
3 years ago
Nate's weekly pay increased from $710 to $770. Which of the following is closest to the
Maksim231197 [3]

Answer:

about 7.2 I think

Step-by-step explanation:

if his pay went up all the way to 770 it would be going up either 7.8 or 7.2

7 0
3 years ago
If a and b are two angles in standard position in Quadrant I, find cos(a+b) for the given function values. sin a=8/17and cos b=1
n200080 [17]

Recall the sum identity for cosine:

cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

so that

cos(a + b) = 12/13 cos(a) - 8/17 sin(b)

Since both a and b terminate in the first quadrant, we know that both cos(a) and sin(b) are positive. Then using the Pythagorean identity,

cos²(a) + sin²(a) = 1   ⇒   cos(a) = √(1 - sin²(a)) = 15/17

cos²(b) + sin²(b) = 1   ⇒   sin(b) = √(1 - cos²(b)) = 5/13

Then

cos(a + b) = 12/13 • 15/17 - 8/17 • 5/13 = 140/221

6 0
2 years ago
Write a system of linear equations for the graph below.
Vanyuwa [196]

Answer:

-3/-2

Step-by-step explanation:

because (x,y) the x coordinate is -3 and the y coordinate is -2

3 0
2 years ago
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