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Free_Kalibri [48]
3 years ago
7

Please answer the question from the attachment.

Mathematics
1 answer:
Nostrana [21]3 years ago
5 0
10-4=6
3*2k= 6k
6k=6
K=1 is the answer
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Your favorite craft store had a weekend sale. On the first day, all $10 items were 70% off. You bought some number of these of t
Svet_ta [14]
For the first day we have the following function:
 f (n) = (0.3 * 10) n + 8
 For the second day we have the following function:
 f (n) = (0.4 * 10) n + 5
 You spent the same amount of money as the day before:
 (0.3 * 10) n + 8 = (0.4 * 10) n + 5
 3n + 8 = 4n + 5
 n = 8-5
 n = 3 items
 We evaluate each function for n = 3
 f (3) = (0.3 * 10) * 3 + 8 = 17 $
 f (3) = (0.4 * 10) * 3 + 5 = 17 $
 The total amount of money is:
 17 + 17 = 34 $
 Answer:
 
you purchase 6 items
 
you spend at the craft store during the sale $ 34
8 0
3 years ago
Angelina paid $6.60 for a dozen cupcakes for a birthday party. What is the unit cost for each cupcake? *
S_A_V [24]

Answer:$.55

Step-by-step explanation:

An important thing to know is that a dozen is equal to 12.

12x= 6.60

/12     /12

x=.55

3 0
3 years ago
Read 2 more answers
Can somebody pls answer these three questions pls!!!!ASAP
Y_Kistochka [10]

Answer:

1) x=1

2) x=6

3) x=3

Step-by-step explanation:

i just did them :) good luck

7 0
3 years ago
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}



4 0
4 years ago
I don't know pls pls plsssssss help me I don't understand!!!!‍♂️
Andre45 [30]
The answer will be 200
7 0
3 years ago
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