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3 years ago

Help plz help plz help plz help plz

Mathematics

2 answers:

NARA [144]3 years ago

8 0

Answer:

Step-by-step explanation:

21. 136° and b are supplementary, meaning they have a sum of 180°. This means that 136° + b = 180° → b = 44°.

22. Again, the angles are supplementary which means that 61° + b = 180° → b = 119°.

23. Again, they are supplementary which means that x - 22° + 135° = 180° → x + 113° = 180° → x = 67°.

24. I'm pretty sure you get the gist of this. 6x + 3° + 87° = 180° → 6x + 90° = 180° → 6x = 90° → x = 15°

Hope this helps!

kow [346]3 years ago

7 0

Answer:

139o

Step-by-step explanation:

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Solve x 3x+6=42 please help me get this done

Nikitich [7]

_______________________________

Solution,

3x+6=42

or,3x=42-6

or, 3x=36

or,X=36/3

X=12

The value of X is 12.

hope it helps

Good luck on your assignment

4 0

3 years ago

Read 2 more answers

Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4%

bogdanovich [222]

Mr. Jackson invested $800 at 6% per year and $ 2400 at 4 % per year

<h3>Solution:</h3>

Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4% per year.

Let the sum invested be ‘a’ and ‘3a’ at 6% per year and 4 % per year respectively

Also, his annual return totaled $144

We can form following equation on the basis of question:-

$\begin{array}{l}{\text { Then, } \frac{a \times 6 \times 1}{100}+\frac{3 a \times 4 \times 1}{100}=\$ 144} \\\\ {\frac{6 a}{100}+\frac{12 a}{100}=144} \\\\ {\frac{6 a+12 a}{100}=144} \\\\ {\frac{18 a}{100}=144} \\\\ {18 a=14400} \\\\ {a=14400 \div 18}\end{array}$

a = $800

The amount of money invested at 6% = a = 800

The amount of money invested at 4 % = 3a = 3(800) = 2400

So, the amount of money invested at 6% is $800 and the amount of money invested at 4% is $ 2400

4 0

3 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

Juliette [100K]

Hello,
Please, see the attached files.
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3 0

3 years ago

Help with q25 please. Thanks.

Westkost [7]

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

$f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\$

Now apply the derivative to that to get the second derivative

$f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\$

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

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

Let's compute dy/dx. We'll use f(x) as defined earlier.

$y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\$

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\$

If you need a refresher on the quotient rule, then

$\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\$

where P and Q are functions of x.

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

This then means

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\$