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3 years ago

Let mA=40 degrees . If B is a complement of A and C is a supplement of B find the measures

2 answers:

6 0

If B is the complement then:

complement angles add up to 90°

90 - 40 = ∠B
50 = ∠B

If C is the complement of B then:

90 - 50 = ∠C
40 = ∠C

Hope this helps :)

s2008m [1.1K]3 years ago

4 0

Hello!

Complementary angles mean they add to 90

Since a and b are complementary we can use the equation

a + b = 90

Put in the values you know

40 + b = 90

Subtract 40 from both sides

b = 50

supplementary means they add to 180

Since B and C are supplementary we can use the equation

B + C = 180

Put in the values you know

50 + c = 180

subtract 180 from both sides

c = 130

The answers are b = 50 and c = 130

Hope this helps!

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What value of b will cause the system to have an infinite number of solutions? A system of equations. y equals 6 x plus b. negat

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The value of b is 6 for which the system of equation will have an infinite number of solutions option third is correct.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

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The complete question is in the picture, please refer to the attached picture.

We have two linear equations:

y = 6x - b ..(1)

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From the second equation:

$\rm -3x + \dfrac{1}{2}y = -3$

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y = 6x - 6 ..(3)

On comparing equation (1) and (3)

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2 years ago

List these slopes in order from least to greatest:<br><br> m= <br> undefined, 2, 0, -3, .5, -2

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2 years ago

A student takes a driving test until it is passed.

Leya [2.2K]

Answer:

The probability that the test is taken an even number of times is 0.30.

Step-by-step explanation:

The probability that a student passes the driving test at any attempt is,

<em>p</em> = 4/7.

The event of a student passing in any attempt is independent of each other.

The probability that the test is taken an even number of times is:

P (even number of tests) = P (Passing in the 2nd attempt)

+ P (Passing in the 4th attempt)

+ P (Passing in the 6th attempt) ...

If a student passed in the 2nd attempt it implies that he failed in the first.

Then, P (Passing in the 2nd attempt) = $(\frac{3}{7}) \times (\frac{4}{7})$

Similarly, P (Passing in the 4th attempt) = $(\frac{3}{7})^{3} \times (\frac{4}{7})$ , since he failed in the first 3 attempts.

And so on.

Compute the probability of an even number of tests as follows:

P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

The result follows a Geometric progression for infinite values.

The sum of infinite GP is:

$S=\frac{a}{1-r^{2}}$

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P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

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3 years ago

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Answer:

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