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3 years ago

One sixth of a number?

Mathematics

1 answer:

rodikova [14]3 years ago

5 0

One-sixth of a number is four

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Help this is really hard

scoundrel [369]

Answer:

x = 18, y = 45

Step-by-step explanation:

4x - 7 + a = 180 (linear pair)

But, a = 6x + 7 (alternate interior angles)

=> 4x - 7 + 6x + 7 = 180

=> 10x = 180

=> x = 18

Now, 3y - 20 = 6x + 7 (vertically opposite angles)

=> 3y - 20 = 6(18) + 7

=> 3y = 108 + 7 + 20

=> 3y = 135

=> y = 45

Hope it helps :)

Please mark my answer as the brainliest

4 0

2 years ago

Read 2 more answers

it rained for 7 days during the month of april . About what percent did it rain ( hint aprill has 30 days ) round to the nearst

oee [108]

Answer:

Convert fraction (ratio) 7 / 30 Answer: 23.333333333333%

Step-by-step explanation:

7 0

2 years ago

How do I calculate f(x) from a graph

slamgirl [31]

Locate the number 2 on the x axis, draw a vertical line to the graph of the function, and then draw a horizontal line to the y axis, as shown below. The coordinates of the point are (2, f(2)). The y value of the point is f(2). Read the y value of this point on the y axis

4 0

3 years ago

Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$