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2 years ago

( Need answered fast please)What is the value of c? c c–35° c–50°

Mathematics

1 answer:

Vlad1618 [11]2 years ago

4 0

Answer:

c = 85°

Step-by-step explanation:

Inside of the triangle angles:

c–35°
c–50°
180°-c

these angle sum up to 180°

180-c+c–35°+c–50° = 180°

c +95° = 180°

c = 180° - 95°

c = 85°

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The ratio of Wei Ling's age to Wei Xuan's age 5 years ago was 2: 5. In 9 years

Archy [21]

Answer:

15 years old

Step-by-step explanation:

Start by defining the variables that we are going to use throughout our working:

Let the current age of Wei Ling and Wei Xuan be L and X years old respectively.

Next, form equations using the given information.

5 years ago

Wei Ling: (L -5) years old

Wei Xuan: (X -5) years old

Given that the ratio of Wei Ling's age to that of Wei Xuan's is 2: 5,

$\frac{ L - 5}{X - 5} = \frac{2}{5}$

Cross multiply:

2(X -5)= 5(L -5)

Expand:

2X -10= 5L -25

2X= 5L -25 +10

2X= 5L -15 -----(1)

9 years time

Wei Ling: (L +9) years old

Wei Xuan: (X +9) years old

Given that the ratio of Wei Ling's age to that of Wei Xuan is 3: 4,

$\frac{L + 9}{X + 9} = \frac{3}{4}$

Cross multiply:

3(X +9)= 4(L +9)

Expand:

3X +27= 4L +36

3X= 4L +36 -27

3X= 4L +9 -----(2)

Let's solve using the elimination method.

(1) ×3:

6X= 15L -45 -----(3)

(2) ×2:

6X= 8L +18 -----(4)

(3) -(4):

6X -6X= 15L -45 -(8L +18)

0= 15L -45 -8L -18

0= 7L -63

7L= 63

L= 63 ÷7

L= 9

Substitute L= 9 into (1):

2X= 5(9) -15

2X= 45 -15

2X= 30

X= 30 ÷2

X= 15

Thus, Wei Xuan is 15 years old now.

6 0

2 years ago

Trọng lượng một sp là bnn pp chuẩn với trung bình là 20,5 và độ lệch chuẩn là 2. Lấy ngẫu nhiên 4 sp để kiểm tra thì với xs 0,95

Illusion [34]

You need to make this is all languages if you want help thank you . :)

8 0

3 years ago

What’s 32% of ____=8

Zanzabum

Answer:

Step-by-step explanation:

32% = 32/100 = 0.32

Let the unknown number be x.

0.32 x = 8

x = 25

6 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

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

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

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

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form ( f(x) )^2 in the numerator to cancel out with the denominator.

$\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\$

So this concludes the proof that $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\$ when $y = \ln\left(\sin(px)+\cos(px)\right)\\\\$

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

7 0

2 years ago

I need the intervals please

serg [7]

Answer:

1. (-6, -5) to (-5, -4)

2. (-3, -4) to (-1, -1)

Step-by-step explanation:

Increasing intervals include the x- and y- points increasing.

PS: I don't really know how to present the answer so if you're gonna submit it for a test or smth pls improve on the formatting yourself based on the question (So sorry I didn't really understand the question about writing a list of intervals and separating them with a comma :P)

5 0

2 years ago