All categories

2 years ago

Help me with this please!!!!!!

Mathematics

1 answer:

castortr0y [4]2 years ago

4 0

Answer:

$\huge\mathcal\red{here's \: your \: solution..} \\ \\ \cos = base \div hypotenuse \: \\ \\ \cos(60) = 1 \div 2 \\ \\ \cos = a \div 14 \\ \\ a \div 14 = 1 \div 2 \\ \\ a = 14 \div2 \\ \\ a = 7 \: cm \\ \\ \huge\mathfrak\green{hope \: it \: helps..}$

You might be interested in

Darius lives in Ohio and pays 5.5% in sales tax. If he just bought a sweatshirt that cost $24, what was the total amount he paid

ollegr [7]

In this question, we are required to find the total amount he paid to buy the shirt.

It is given that sales tax is 5.5 % over the amount paid.

The cost price of the shirt = $ 24.

The total amount will be = Cost price + 5.5 % of Cost price

Total amount paid = $ 24 × 1.055 (i.e. 100 % + 5.5 %)

Total amount = $ 24 + (5.5 % × $ 24)

Total amount paid = $ 24 + $ 1.32 = $ 25.32

8 0

2 years ago

Read 2 more answers

A circular mirror has a circumference of 30 pi inches. What is the area of mirror?

Aleonysh [2.5K]

Answer:

Step-by-step explanation:

circumference = 2πr

2πr = 30π inches

$r=\frac{30\pi}{2\pi}=15in$

Area = πr² = π*15*15 = 225π square inches

6 0

2 years ago

8) through: (-3, -2), perp. to y = x – 1 A) y=-5x – 1 B) y=-4x – 5 C) y=-x – 5 D) y=-5x – 4

nexus9112 [7]

Answer:

The equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

Solution:

Given, line equation is y = x – 1 ⇒ x – y – 1 = 0. And a point is (-3, -2)

We have to find the line equation which is perpendicular to above given line and passing through the given point.

Now, let us find the slope of the given line equation.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-1}=1$

We know that, product of slopes of perpendicular lines is -1.

So, 1 $\times$ slope of perpendicular line = -1

slope of perpendicular line = -1

Now let us write point slope form for our required line.

$\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

y – (-2) = -1(x – (-3))

y + 2 = -1(x + 3)

y + 2 = -x – 3

x + y + 2 + 3 = 0

x + y + 5 = 0

y = -x -5

Hence the equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

8 0

3 years ago

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$