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If m∠ABC = 72°, what is m∠DAB?
HELP If m∠ABC = 72°, what is m∠DAB? - 1

Answers

Answer 1
Answer:

Value of ∠DAB is 108°

Intersection of Parallel lines:

Given that;

∠ABC = 72°

Find:

Value of ∠DAB

Computation:

We know that Line AD is parallel to line CB

So,

∠ABC + ∠DAB = 180°

72° + ∠DAB = 180°

∠DAB = 108°

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Answer 2
Answer:

Answer: m∠DAB=72

Step-by-step explanation:

Since we have given that

∠ABC = 72°

Since the lines AD and BC are parallel to each other,

As we know that when the two parallel lines cut by a transversal then the corresponding angles will  be formed.

And the corresponding angles will be equal.

So, m∠ABC =m∠DAB=72°

Hence, m∠DAB=72°



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Find the equation of this line.

Answers

f(x)=-2/5x+1

using the two dots Rise/run =-4/10 simplifies tp -2/5 y-intercept is 1

Help please help with this question

Answers

first you have to do puthagorean theorm to find the missing leg which is QR then to find sinS you go opposite over hypotenuse. simplify if needed.
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What is the 44th decimal digit in the decimal representation of 1/11?

Answers

The decimal representation of  1/11  is  0.090909...
The pair of digits '09' repeat forever.

If you start counting the digits after the decimal point,
the odd ones are all zero, and the even ones are all  9 .

So the 44th digit after the decimal point is  9 .


Final answer:

The 44th decimal digit in the decimal representation of 1/11 is 9, as the pattern '09' repeats indefinitely.

Explanation:

The decimal representation of 1/11 is 0.090909..., which is a repeating decimal. The pattern '09' repeats indefinitely. To find the 44th decimal digit, we can divide 44 by 2 (since each '09' is 2 digits), which equals 22 with remainder 0. This means that the 44th digit is the second digit of the 22nd '09' pair, so the 44th decimal digit in the decimal representation of 1/11 is 9.

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A, B & C form a triangle where ∠ BAC = 90°. AB = 13.5 mm and CA = 3.2 mm. Find the length of BC, giving your answer rounded to 1 DP.

Answers

Answer:

13.9

Step-by-step explanation:

Use Pythagorean Theorem:

Both of the lengths given are the legs,

13.5^2 + 3.2^2 = c^2

182.25 + 10.24 = c^2

192.49 = c^2

13.87 = c

3.30 Measurements of scientific systems are always subject to variation, some more than others. There are many structures for measurement error, and statisticians spend a great deal of time modeling these errors. Suppose the measurement error X of a certain physical quantity is decided by the density function f(x) = k(3 − x2), −1 ≤ x ≤ 1, 0, elsewhere. (a) Determine k that renders f(x) a valid density function. (b) Find the probability that a random error in measurement is less than 1/2. (c) For this particular measurement, it is undesirable if the magnitude of the error (i.e., |x|) exceeds 0.8. What is the probability that this occurs?

Answers

Answer:

a) k should be equal to 3/16 in order for f to be a density function.

b) The probability that the measurement of a random error is less than 1/2 is 0.7734

c) The probability that the magnitude of a random error is more than 0.8 is 0.164

Step-by-step explanation:

a) In order to find k we need to integrate f between -1 and 1 and equalize the result to 1, so that f is a density function.

1 = k \int\limits^1_(-1) {(3-x^2)} \, dx = k  *  (3x-(x^3)/(3))|_(x=-1)^(x = 1) = k*[(3-1/3) - (-3 + 1/3)] = 16k/3

16k/3 = 1

k = 3/16

b) For this probability we have to integrate f between -1 and 0.5 (since f takes the value 0 for lower values than -1)

P(X < 1/2) = \int\limits^(0.5)_(-1) {(3)/(16)(3-x^2)} \, dx = (3)/(16) [(3x-(x^3)/(3)) |_(x=-1)^(x=0.5)] =(3)/(16) *(1.458333 - (-3+1/3)) = 0.7734

c) For |x| to be greater than 0.8, either x>0.8 or x < -0.8. We should integrate f between 0.8 and 1, because we want values greater than 0.8, and f is 0 after 1; and between -1 and 0.8.

P(|X| > 0.8) = \int\limits^(-0.8)_(-1) {(3)/(16)*(3-x^2)} \, dx + \int\limits^(1)_(0.8) {(3)/(16)*(3-x^2)} \, dx =\n (3)/(16) (3x-(x^3)/(3))|_(x=-1)^(x=-0.8) + (3)/(16) (3x-(x^3)/(3))|_(x=0.8)^(x=1) = 0.082 + 0.082 = 0.164

(a) The value of k that makes f(x) a valid density function is k = 1/6.

(b) The probability that a random error in measurement is less than 1/2 is 3/4.

(c) The probability that the magnitude of the error exceeds 0.8 is 1/4.

(a) To make the given function f(x) a valid probability density function, it must satisfy the following conditions:

The function must be non-negative for all x: f(x) ≥ 0.

The total area under the probability density function must equal 1: ∫f(x)dx from -1 to 1 = 1.

Given f(x) = k(3 - x^2), -1 ≤ x ≤ 1, and f(x) = 0 elsewhere, let's find the value of k that satisfies these conditions.

Non-negativity: The function is non-negative for -1 ≤ x ≤ 1, so we have k(3 - x^2) ≥ 0 for -1 ≤ x ≤ 1. This means that k can be any positive constant.

Total area under the probability density function: To find the value of k, integrate f(x) over the interval [-1, 1] and set it equal to 1:

∫[from -1 to 1] k(3 - x^2)dx = 1

∫[-1, 1] (3k - kx^2)dx = 1

Now, integrate the function:

[3kx - (kx^3/3)] from -1 to 1 = 1

[(3k(1) - (k(1^3)/3)) - (3k(-1) - (k(-1^3)/3))] = 1

Simplify:

[3k - k/3 + 3k + k/3] = 1

6k = 1

k = 1/6

So, the value of k that makes f(x) a valid density function is k = 1/6.

(b) To find the probability that a random error in measurement is less than 1/2, you need to calculate the integral of f(x) from -1/2 to 1/2:

P(-1/2 ≤ X ≤ 1/2) = ∫[from -1/2 to 1/2] f(x)dx

P(-1/2 ≤ X ≤ 1/2) = ∫[-1/2, 1/2] (1/6)(3 - x^2)dx

Now, integrate the function:

(1/6) [3x - (x^3/3)]from -1/2 to 1/2

[(1/6)(3(1/2) - ((1/2)^3/3)) - (1/6)(3(-1/2) - ((-1/2)^3/3))]

Simplify:

(1/6)[(3/2 - 1/24) - (-3/2 + 1/24)]

(1/6)[(9/8) + (9/8)]

(1/6)(18/8)

(3/4)

So, the probability that a randomerror in measurement is less than 1/2 is 3/4.

(c) To find the probability that the magnitude of theerror (|x|) exceeds 0.8, you need to calculate the probability that |X| > 0.8. This is the complement of the probability that |X| ≤ 0.8, which you can calculate as:

P(|X| > 0.8) = 1 - P(|X| ≤ 0.8)

P(|X| > 0.8) = 1 - P(-0.8 ≤ X ≤ 0.8)

We already found P(-0.8 ≤ X ≤ 0.8) in part (b) to be 3/4, so:

P(|X| > 0.8) = 1 - 3/4

P(|X| > 0.8) = 1/4

So, the probability that the magnitude of the error exceeds 0.8 is 1/4.

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Does the graph represent an exponential growth
or an exponential decay function?

Answers

there’s no graph......

Answer:

Ummmm... I'm sorry but there is no graph or equation, so I can't answer this question with them...