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Masja [62]
3 years ago
13

Please help...given that elm and oak street are parallel, find m

Mathematics
2 answers:
vova2212 [387]3 years ago
7 0

Answer:

∠a = 3x -12 ( Being corresponding angle)

Now

(3x-12) + (2x + 2) = 180 ( Sum of straight angles)

5x - 10 = 180

5x = 190

x = 38

m∠a = 3x -12 = 3*38 -12 = 102

Step-by-step explanation:

jarptica [38.1K]3 years ago
6 0

Answer:

<em>102°</em>

Step-by-step explanation:

(3x - 12)° + (2x + 2)° = 180°

5x - 10 = 180 ⇒ x = 38

m∠A = (3x - 12)°

<em>m∠A</em> = (3×38 - 12)° = <em>102°</em>

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Which factorization is incorrect? 1) x3 + 5x2 + 4x + x2 + 5x + 4 = (x + 4)(x + 1)2 2) 4d3 – 4d2 – 9d + 9 = (d – 1)(2d – 3)2 3) 2
Andru [333]

Answer:

The second option is the incorrect:

2) 4d3 – 4d2 – 9d + 9 = (d – 1)(2d – 3)^2

Step-by-step explanation:

Let's check each one of the factorizations:

1) x3 + 5x2 + 4x + x2 + 5x + 4 = (x + 4)(x + 1)^2

(x + 4)(x + 1)^2 = (x + 4)(x2 + 2x + 1) = x3 + 6x2 + 9x + 4

x3 + 5x2 + 4x + x2 + 5x + 4 = x3 + 6x2 + 9x + 4

This one is correct.

2) 4d3 – 4d2 – 9d + 9 = (d – 1)(2d – 3)^2

(d – 1)(2d – 3)^2 = (d - 1)(4d2 - 12d + 9) = 4d3 - 16d2 + 21d - 9

4d3 – 4d2 – 9d + 9 = 4d3 - 16d2 + 21d - 9

This one is incorrect.

3) 27h3 – 8k3 = (3h – 2k)(9h2 + 6hk + 4k2 )

(3h – 2k)(9h2 + 6hk + 4k2 ) = 27h3 - 8k3

This one is correct.

4) 64 – 9t2 = (8 + 3t)(8 – 3t)

(8 + 3t)(8 – 3t) = 64 - 9t2

This one is correct.

So the incorrect option is the second one.

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3 years ago
At a certain university, the average attendance at basketball games has been 3125. Due to the dismal showing of the team this ye
Neporo4naja [7]

Answer:

The test statistic needed to evaluate the claim is t = -1.08.

Step-by-step explanation:

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the expected value of the mean, s is the standard deviation of the sample and n is the size of the sample.

At a certain university, the average attendance at basketball games has been 3125. The athletic director claims that the attendance is the same as last year.

This means that \mu = 3125

Due to the dismal showing of the team this year, the attendance for the first 9 games has averaged only 2915 with a standard deviation of 585.

This means that n = 9, X = 2915, s = 585

What is the test statistic needed to evaluate the claim?

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{2915 - 3125}{\frac{585}{\sqrt{9}}}

t = -1.08

The test statistic needed to evaluate the claim is t = -1.08.

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