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3 years ago

Parallel lines e and f are cut by transversal b. What is the value of x? A.130 B.16 C.164 D.50

Mathematics

1 answer:

Gnesinka [82]3 years ago

4 0

Answer:

A) 130

Step-by-step explanation:

(2x + 18) + y = 180 <--- Is True

(4x - 14) + y = 180 <--- Is True

(2x + 18) + y = y + (4x - 14)

(2x + 18) = (4x - 14)

2x + 18 = 4x - 14

-2x + 18 = -14

- 18 - 18

-2x = -32

x = 16

2(16) + 18 = 50

4(16) - 14 = 50

180 - 50 = 130

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A triangle is to be dilated with the origin as the center of dilation and with a scale factor of 2. Which answer choice correctl

saveliy_v [14]

Answer:

All the coordinates are changed to

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Step-by-step explanation:

Given - A triangle is to be dilated with the origin as the center of dilation and with a scale factor of 2.

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Proof -

Let ABC is a triangle where A, B, C are the vertices of the triangle respectively.

Now,

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

What is the height of a triangular prism that has a height of 9 meters and a base with the following deminsions 4m 14m

Andrew [12]

Answer:

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Step-by-step explanation:

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3 years ago

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Ksenya-84 [330]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

In the 1990s the demand for personal computers in the home went up with household income. For a given community in the 1990s, th

WITCHER [35]

Answer:

a) 0.5198 computers per household

b) 0.01153 computers

Step-by-step explanation:

Given:

number of computers in a home,

q = 0.3458 ln x - 3.045 ; 10,000 ≤ x ≤ 125,000

here x is mean household income

mean income = $30,000

increasing rate, $\frac{dx}{dt}$ = $1,000

Now,

a) computers per household are

since,

mean income of $30,000 lies in the range of 10,000 ≤ x ≤ 125,000

thus,

q = 0.3458 ln(30,000) - 3.045

q = 0.5198 computers per household

b) Rate of increase in computers i.e $\frac{dq}{dt}$

$\frac{dq}{dt}$ = $\frac{d(0.3458 ln x - 3.045)}{dt}$

$\frac{dq}{dt}=0.3458\times(\frac{1}{x})\frac{dx}{dt} - 0$

on substituting the values, we get

$\frac{dq}{dt}=0.3458\times(\frac{1}{30,000})\times1,000$

= 0.01153 computers

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3 years ago