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3 years ago

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kipiarov [429]3 years ago

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Find the area of the circle. Leave your answer in terms of pi

puteri [66]

Answer:

Step-by-step explanation:

Since area of a circle is pi.r^2

We know that diameter is 20

Radius is half of diameter

So it becomes

Pi.(d/2)^2

So pi.(40/4) is the area of this circle.

Or 40pi/4

3 0

3 years ago

Read 2 more answers

Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

Rainbow [258]

Answer:

$\sin L = 0.60$

$tan\ N = 1.33$

$\cos L = 0.80$

$\sin N = 0.80$

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

7 0

3 years ago

Solve for Y:<br> 2/3 (y+57 )= 178

sattari [20]

Y= 210 you have to distribute 2/3 to y then to 57 and your equation would be 2/3y + 38= 178 then you take 38 and subtract 178 - 38 and you get 140 then you divide 140/ (2/3) and you will get 210 as your answer.(:

8 0

3 years ago

In a normal distribution what percent of values fall within one standard deviation of the mean?

Sophie [7]

The answer is given by the empirical rule, 68%.

8 0

3 years ago

Plz i need help and plz explain

Montano1993 [528]

$\frac{ {x}^{2} - 7x + 12}{ {x}^{2} - x - 12 } \\ = \frac{ {x}^{2} - 4x - 3x + 12 }{ {x}^{2} - 4x + 3x - 12 } \\ = \frac{x(x - 4) - 3(x - 4)}{x(x - 4) + 3(x - 4)} \\ = \frac{(x - 3)(x - 4)}{(x + 3)(x - 4)} \\ = \frac{x - 3}{x + 3}$

x² - x - 12 ≠ 0

x² - 4x + 3x - 12 ≠ 0

x (x - 4) + 3 (x - 4) ≠ 0

(x + 3)(x - 4) ≠ 0

x ≠ -3 or x ≠ 4

(B)

3 0

3 years ago