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3 years ago

Find the cos sin and tan

Mathematics

1 answer:

Taya2010 [7]3 years ago

3 0

Answer:

Step-by-step explanation:

Sin= o/h = 23/40 = 0.575 = sin^-1(0.575) = 35.1 degrees

Cos = a/h = 37/40 = 0.925 = cos^-1(0.925) = 22.3 degrees

Tan= o/a = 23/37 = 0.622 = tan^-1(0.622) = 31.9 degrees

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13x = 79 ? Can somebody explain #9 b. PLEASE. Please solve using the system of linear equations by elimination.I would love it i

strojnjashka [21]

8x + 2y = 44
5x + 2y = 35

Since 2y is already common, there is no need for multiplication. Just reverse the signs on the equation.

Like this;

8x + 2y = 44
-5x - 2y = -35

All the additions become subtractions and vice versa.

-2y and +2y get cancelled.

Leaving;

8x = 44
-5x = -35

Now, subtract again.

8x - 5x = 3x

44 - 35 = 9

So;

3x = 9

x = 9 / 3

x = 3

We found 'x' and now we have to find 'y'.

Just substitute.

[8 x 3] + 2y = 44

24 + 2y = 44

2y = 44 - 24

2y = 20

y = 20 / 2

y = 10

Now, we re-check;

[8 x 3] + [2 x 10] = 44

24 + 20 = 44

Explaining the error for #9b is simply doing the whole sum.

Hope this helped! :) I tried my best to explain.

7 0

3 years ago

Find the limit of the function by using direct substitution.

serg [7]

Answer:

Option a.

$\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1$

Step-by-step explanation:

You have the following limit:

$\lim_{x \to \frac{\pi}{2}{(3e)^{xcosx}$

The method of direct substitution consists of substituting the value of $\frac{\pi}{2}$ in the function and simplifying the expression obtained.

We then use this method to solve the limit by doing $x=\frac{\pi}{2}$

Therefore:

$\lim_{x \to \frac{\pi}{2}}{(3e)^{xcosx} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})}$

$cos(\frac{\pi}{2})=0\\$

By definition, any number raised to exponent 0 is equal to 1

$\lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}(0)}\\\\$

$\lim_{x\to \frac{\pi}{2}}{(3e)^{0}} = 1$

Finally

$\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1$

6 0

3 years ago

How do you compare two-way frequencies in a relative frequency table

Nikolay [14]

To avoid such problems when comparing the categorical variables in a two-way frequency table, we need to exam the table by separate categories (rows or columns). When a relative frequency is determined based upon a row or column, it is called a "conditional" relative frequency.
Two-way relative frequency tables show what percent of data points fit in each category. ...
For example, here's how we would make column relative frequencies:
Step 1: Find the totals for each column.
Step 2: Divide each cell count by its column total and convert to a percentage.

3 0

3 years ago

Solve each equation using inverse operations: 1. 9w = -54 2. b - 12 = 3 3. n/4 = -11

katovenus [111]

Hello!

Answer:

1. $\boxed{W=-6}$

*The answer must have A NEGATIVE SIGN ONLY!*

2. $\boxed{B=15}$

*The answer must have A POSITIVE SIGN ONLY!*

3. $\boxed{N=-44}$

*The answer must have a NEGATIVE SIGN ONLY!*

Step-by-step explanation:

1. 9w=-54

First, you divide by 9 both sides of an equation.

9w/9=-54/9

Simplify.

-54/9=-6

W=-6 is the final answer.

____________________________________

2. b-12=3

First you add by 12 both sides of an equation.

b-12+12=3+12

Add by the numbers from left/right.

3+12=15

b=15 is the final answer.

__________________________________________

3. n/4=-11

First you multiply by 4 both sides of an equation.

4n/4=4(-11)

Multiply by the numbers from left/right.

4*-11=-44

n=-44 is the final answer.

_____________________________

Hope this helps you!

Have a great day! :)

-Charlie

Thanks!

3 0

2 years ago

In the equation (x^2+y)^5, what is the coefficient of the term x^4y^3? what is the coefficient of the same term in the expansion

lys-0071 [83]

$\displaystyle
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k$

-------------------------------------------------------------

$\displaystyle
(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\
n=5\\
k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10$

It's 10.

----------------------------------------------------

$\displaystyle
(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\
n=5\\
k=4\\\\
\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90$

It's 90

6 0

3 years ago