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3 years ago

15

It costs $22.50 to enter an amusement park and $0.25 to ride a ride. You have $23.75. Write an equation that represents the numb

er r of rides you can ride.

1 answer:

AfilCa [17]3 years ago

3 0

Answer:

Step-by-step explanation:

y = mx + c

y = c + mx

23.75 = 22.50 + 0.25(x)

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In rectangle MNOP, MO = 4x - 10 and NP = 2x + 4.<br> What is the length of a diagonal?

slega [8]

Answer:

18

Step-by-step explanation:

From the rectangle, M O = NP

4x - 10 = 2x+4

4x-2x = 4 + 10

2x = 14

x = 14/2

x = 7

SInce MO and NP are the diagonal

Length of diagonal = 4x- 10

Length of diagonal = 4(7) - 10

Length of diagonal = 28 - 10

Length of diagonal = 18

hence the required length is 18

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3 years ago

There are 150 trees in the park. Elm trees represent 40% of the trees in the park.

deff fn [24]

Answer:

375 trees, hope this help!

5 0

2 years ago

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How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

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3 years ago

What’s 5.9876 rounded to the nearest tenth? (In the picture it’s question #1)

sammy [17]

Its six because its bigger than 5

6 0

3 years ago

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In what time will $400 amount to $512 if the simple interest is calculated at 14% per annum?

kupik [55]

$512\$-400\$=112\$\\\\\frac{112\$}{400\$}\cdot100\%=0,28\cdot100\%=28\%\\\\28:14=2-years$

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3 years ago

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