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2 years ago

9

In an isosceles triangle, the angle formed by its legs is called__

2 answers:

Alecsey [184]2 years ago

7 0

Answer:

a:vertex angle

Step-by-step explanation:

it is vertex angle

nasty-shy [4]2 years ago

7 0

$\large\bf\blue{answer}$

In an isosceles triangle, the angle formed by its legs is called vertex angle

<u>option </u><u>a </u><u>is </u><u>correct </u>

<h3>i hope it helped u</h3>

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Need some great assistance! Thanks!

Lostsunrise [7]

Answer:

b

Step-by-step explanation:

step by step explanation I'm probly wrong

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2 years ago

Graph the system of equations on your graph paper to answer the question. {y=−x+3

makkiz [27]

The answer is (x, y) = (-1, 4) hope this helped

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3 years ago

Jillian walked 0.5 miles before she started jogging at an average pace of 5 miles per hour. The equation d = 0.5 + 5t can be use

Mekhanik [1.2K]

Answer: independent variable: time (t). Dependent variable: distance (d).

Step-by-step explanation: an independent variable is a variable whose variation does not depend on that of another. In the given problem, the independent variable is time, because time will pass by no matter what (without depending in any other variable). The dependent variable in this case is the distance, because how far Jillian goes, depends on how much time she expends walking and jogging.

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3 years ago

Read 2 more answers

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

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2 years ago

((sinB)/1+cosB) + ((1+cosB)/sinB) = 2cscB prove that the equation is an identity

Contact [7]

Step-by-step explanation:

If you need any explanation, we can communicate normally

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2 years ago