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3 years ago

Four runners are training for long races. Noah Ran 5.123 miles. Andre ran 6.34 miles. Jada ran 7.1 miles. Diego ran 8 miles.

Mathematics

1 answer:

Korolek [52]3 years ago

3 0

Jada ran 1.977 miles more than Noah

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Choose the correct simplification of the quantity of x to the 7th power over y to the 9th power all raised to the 2nd power.

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X to the 14th power over y to the 18th power is ur answer

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3 years ago

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Please help me with the below question.

Alexus [3.1K]

a) Substitute $y=x^9$ and $dy=9x^8\,dx$ :

$\displaystyle \int x^8 \cos(x^9) \, dx = \frac19 \int 9x^8 \cos(x^9) \, dx \\\\ = \frac19 \int \cos(y) \, dy \\\\ = \frac19 \sin(y) + C \\\\ = \boxed{\frac19 \sin(x^9) + C}$

b) Integrate by parts:

$\displaystyle \int u\,dv = uv - \int v \, du$

Take $u = \ln(x)$ and $dv=\frac{dx}{x^7}$ , so that $du=\frac{dx}x$ and $v=-\frac1{6x^6}$ :

$\displaystyle \int \frac{\ln(x)}{x^7} \, dx = -\frac{\ln(x)}{6x^6} + \frac16 \int \frac{dx}{x^7} \\\\ = -\frac{\ln(x)}{6x^6} + \frac1{36x^6} + C \\\\ = \boxed{-\frac{6\ln(x) + 1}{36x^6} + C}$

c) Substitute $y=\sqrt{x+1}$ , so that $x = y^2-1$ and $dx=2y\,dy$ :

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \frac12 \int 2y e^y \, dy = \int y e^y \, dy$

Integrate by parts with $u=y$ and $dv=e^y\,dy$ , so $du=dy$ and $v=e^y$ :

$\displaystyle \int ye^y \, dy = ye^y - \int e^y \, dy = ye^y - e^y + C = (y-1)e^y + C$

Then

$\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \boxed{\left(\sqrt{x+1}-1\right) e^{\sqrt{x+1}} + C}$

d) Integrate by parts with $u=\sin(\pi x)$ and $dv=e^x\,dx$ , so $du=\pi\cos(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \sin(\pi x) \, e^x \, dx = \sin(\pi x) \, e^x - \pi \int \cos(\pi x) \, e^x \, dx$

By the fundamental theorem of calculus,

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = - \pi \int_0^1 \cos(\pi x) \, e^x \, dx$

Integrate by parts again, this time with $u=\cos(\pi x)$ and $dv=e^x\,dx$ , so $du=-\pi\sin(\pi x)\,dx$ and $v=e^x$ :

$\displaystyle \int \cos(\pi x) \, e^x \, dx = \cos(\pi x) \, e^x + \pi \int \sin(\pi x) \, e^x \, dx$

By the FTC,

$\displaystyle \int_0^1 \cos(\pi x) \, e^x \, dx = e\cos(\pi) - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx$

Then

$\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = -\pi \left(-e - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx\right) \\\\ \implies (1+\pi^2) \int_0^1 \sin(\pi x) \, e^x \, dx = 1 + e \\\\ \implies \int_0^1 \sin(\pi x) \, e^x \, dx = \boxed{\frac{\pi (1+e)}{1 + \pi^2}}$

e) Expand the integrand as

$\dfrac{x^2}{x+1} = \dfrac{(x^2 + 2x + 1) - (2x+1)}{x+1} = \dfrac{(x+1)^2 - 2 (x+1) + 1}{x+1} \\\\ = x - 1 + \dfrac1{x+1}$

Then by the FTC,

$\displaystyle \int_0^1 \frac{x^2}{x+1} \, dx = \int_0^1 \left(x - 1 + \frac1{x+1}\right) \, dx \\\\ = \left(\frac{x^2}2 - x + \ln|x+1|\right)\bigg|_0^1 \\\\ = \left(\frac12-1+\ln(2)\right) - (0-0+\ln(1)) = \boxed{\ln(2) - \frac12}$

f) Substitute $e^{7x} = \tan(y)$ , so $7e^{7x} \, dx = \sec^2(y) \, dy$ :

$\displaystyle \int \frac{e^{7x}}{e^{14x} + 1} \, dx = \frac17 \int \frac{\sec^2(y)}{\tan^2(y) + 1} \, dy \\\\ = \frac17 \int \frac{\sec^2(y)}{\sec^2(y)} \, dy \\\\ = \frac17 \int dy \\\\ = \frac y7 + C \\\\ = \boxed{\frac17 \tan^{-1}\left(e^{7x}\right) + C}$

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2 years ago

1) A train travels 25 miles in 1/5 of an hour whats the trains speed in miles per hour?

Radda [10]

Answer:

<em>Hope this helps! ↓</em>

Step-by-step explanation:

1.) 90 Km/h. Example: What is the speed of a cheetah that travels 112.0 meters in 4.0 seconds? ... Speed of the cheetah in miles per hour 28m 1 mile 3600 s 63 miles

2.) y=kx is the direct variation formula. Inserting the information and variables from this scenario we get h = kn.

12 = k*(10) First solve for 'k', the constant of proportionality by dividing by 10.

1.2 = k

Next write the equation using this k and the h & n.

h = 1.2*n

Now we want to find the height if we have 24 books so plug 24 in for n.

h = 1.2 * 24

h = 28.8 inches

3.) 10/3

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3 years ago

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Franklin school is having a Popsicle party for its nine hundred fifty four students.The secretary there will some Popsicle left

mr Goodwill [35]

Answer:

175 boxes

Step-by-step explanation:

There are a total of 954 students since each student will get two popsicles that would mean we need a total of 1,908 popsicles (954 * 2). Since each box contains 12 popsicles then we simply divide the total number of popsicles needed by the number of popsicles that fit in each box to find the total number of boxes.

1,908 / 12 = 159 boxes

Since there are 25 boxes per carton we divide 159 by 25

159 / 25 = 6.36

Now we see that we need at least 6.36 cartons of popsicles, but since we can't get 6.36 we would need to get 7 cartons which would bring a total of

7 * 25 = 175 boxes

Finally, we would end up with 175 boxes of popsicles which would be enough for all the students and have leftover for the teachers.

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3 years ago

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mars1129 [50]

Answer:

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