Ask question

Ask question

All categories

3 years ago

15

Angela wrote the number 186,425 on the board in which number is the value of the digit six exactly 10 times the value of the dig

it six in the Number Andrew wrote

1 answer:

Butoxors [25]3 years ago

4 0

So what’s the question? What are we trying to find?

You might be interested in

A roller coaster at XTRA Fun Rides goes into an underground tunnel during the first descent of the

gregori [183]

Answer:

y-intercept: (0, 75) ⇒ y = 75

x-intercepts: (3, 0) and (5, 0) ⇒ x = 3 and x = 5

vertex: (4, -5)

Step-by-step explanation:

Let the x-axis be the ground (when y = 0). Therefore underground when y < 0 and overground when y > 0

The function is a quadratic: f(x) = ax² + bx + c

If the coaster enters the tunnel three seconds after the first descent starts and stays in the tunnel for a total of 2 seconds, then y = 0 when x = 3 and x = 5

Also, when x = 1, y = 40

f(3) = 0
⇒ 9a + 3b + c = 0

f(5) = 0
⇒ 25a + 5b + c = 0

f(1) = 40
⇒ a + b + c = 40

Solve 2 equations simultaneously:

9a + 3b + c = 0
25a + 5b + c = 0
a + b + c = 40

⇒ a = 5 b = -40 c = 75

⇒ f(x) = 5x² - 40x + 75

<u>y-intercept</u>

when x = 0

⇒ f(0) = 5(0²) - 40(0) + 75 = 75

⇒ y-intercept = (0, 75)

<u>x-intercepts</u>

x = 3 and x = 5

<u>Vertex</u>

If the x-intercepts are when x = 3 and x = 5, then vertex is when x = 4

⇒ f(4) = 5(4²) - 40(4) + 75 = -5

⇒ vertex = (4, -5)

5 0

2 years ago

Calculus, question 5 to 5a

Llana [10]

5. Let $x = \sin(\theta)$ . Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

$\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}$

and $dx = \cos(\theta) \, d\theta$ . So the integral transforms to

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta$

Reduce the power by writing

$\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))$

Now let $y = \cos(\theta)$ , so that $dy = -\sin(\theta) \, d\theta$ . Then

$\displaystyle \int \sin(\theta) (1-\cos^2(\theta)) \, d\theta = - \int (1-y^2) \, dy = -y + \frac13 y^3 + C$

Replace the variable to get the antiderivative back in terms of x and we have

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} + \frac13 \left(\sqrt{1-x^2}\right)^3 + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac13 \sqrt{1-x^2} \left(3 - \left(\sqrt{1-x^2}\right)^2\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \boxed{-\frac13 \sqrt{1-x^2} (2+x^2) + C}$

6. Let $x = 3\tan(\theta)$ and $dx=3\sec^2(\theta)\,d\theta$ . It follows that

$\cos(\theta) = \dfrac1{\sec(\theta)} = \dfrac1{\sqrt{1+\tan^2(\theta)}} = \dfrac3{\sqrt{9+x^2}}$

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

Now,

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \int \frac{27\tan^3(\theta)}{\sqrt{9+9\tan^2(\theta)}} 3\sec^2(\theta) \, d\theta = 27 \int \frac{\tan^3(\theta) \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} \, d\theta$

The denominator reduces to

$\sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)$

and so

$\displaystyle 27 \int \tan^3(\theta) \sec(\theta) \, d\theta = 27 \int \frac{\sin^3(\theta)}{\cos^4(\theta)} \, d\theta$

Rewrite sin³(θ) just like before,

$\displaystyle 27 \int \frac{\sin(\theta) (1-\cos^2(\theta))}{\cos^4(\theta)} \, d\theta$

and substitute $y=\cos(\theta)$ again to get

$\displaystyle -27 \int \frac{1-y^2}{y^4} \, dy = 27 \int \left(\frac1{y^2} - \frac1{y^4}\right) \, dy = 27 \left(\frac1{3y^3} - \frac1y\right) + C$

Put everything back in terms of x :

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac1{\cos^3(\theta)} - \frac3{\cos(\theta)}\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac{\left(\sqrt{9+x^2}\right)^3}{27} - \sqrt{9+x^2}\right) + C$

$\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \boxed{\frac13 \sqrt{9+x^2} (x^2 - 18) + C}$

2(b). For some constants a, b, c, and d, we have

$\dfrac1{x^2+x^4} = \dfrac1{x^2(1+x^2)} = \boxed{\dfrac ax + \dfrac b{x^2} + \dfrac{cx+d}{x^2+1}}$

3(a). For some constants a, b, and c,

$\dfrac{x^2+4}{x^3-3x^2+2x} = \dfrac{x^2+4}{x(x-1)(x-2)} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac c{x-2}}$

5(a). For some constants a-f,

$\dfrac{x^5+1}{(x^2-x)(x^4+2x^2+1)} = \dfrac{x^5+1}{x(x-1)(x+1)(x^2+1)^2} \\\\ = \dfrac{x^4 - x^3 + x^2 - x + 1}{x(x-1)(x^2+1)^2} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac{cx+d}{x^2+1} + \dfrac{ex+f}{(x^2+1)^2}}$

where we use the sum-of-5th-powers identity,

$a^5 + b^5 = (a+b) (a^4-a^3b+a^2b^2-ab^3+b^4)$

4 0

2 years ago

What is the ratio 15 to 35 written as a fraction in lowest terms?

dmitriy555 [2]

B 3/7 because 15 divided by 5 = 3 and 35 divided by 5 = 7

4 0

3 years ago

Read 2 more answers

Please help due asap

alexandr1967 [171]

Answer:

I think im too late- if im not please imform me ill answer- if i get it

Step-by-step explanation:

7 0

2 years ago

The product of the reaction between ethanal and ammonia are what.

kenny6666 [7]

Answer:

hi

Step-by-step explanation:

4 0

3 years ago