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3 years ago

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On Wednesday 34 of the customers who bought gas at a gas station made additional purchases. There were 120 customers who bought

gas. How many of these 120 customers also made additional purchases?

1 answer:

avanturin [10]3 years ago

5 0

Answer:

86

Step-by-step explanation:

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Write the function that shows a translation of f(x) two units to the left.

Galina-37 [17]

Answer:

Step-by-step explanation:

Suppose the base formula is y = x^2

You want to go two units left.

basically that is y = (x + 2)^2 just the opposite of what you might think it should be.

Where does the base function have a minimum? It's minimum is as x = 0 and y = 0

Where does y = (x + 2)^2 have it's minimum?

(-2,0)

conclusion. The base function has moved 2 units to the left.

3 0

3 years ago

At the baseball stadium, the price for popcorn is $5.88 for 3 bags. If you want to buy 7 bags of popcorn, how much will it cost?

4vir4ik [10]

The answer will be $13.72 for 7 bags of popcorn.

to find this answer, you first take 5.88 and divide it by 3 to find out how much one bag costs.

the answer to 5.88/3 is 1.96, so you multiply 1.96 by 7 to get the answer of how much 7 bags of popcorn will cost.

hope this helps!

4 0

3 years ago

Read 2 more answers

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

3 years ago

I am Lyosha [343]

Answer:

see explanation

Step-by-step explanation:

(a)

x² - 2x - 4

using the method of completing the square

add/subtract ( half the coefficient of the x- term )² to x² - 2x

= x² + 2(- 1)x + 1 - 1 - 4

= (x - 1)² - 5

with c = - 1 and d = - 5

(b)

x² - 2x - 4 = 0 ← using result from (a) , that is

(x - 1)² - 5 = 0 ( add 5 to both sides )

(x - 1)² = 5 ( take the square root of both sides )

x - 1 = ± $\sqrt{5}$ ( add 1 to both sides )

x = 1 ± $\sqrt{5}$

5 0

2 years ago

Is the sum of two negative numbers positive or negative

a_sh-v [17]

No, the sum of two negative numbers<span> can never be </span><span>positive</span>

3 0

3 years ago

Read 2 more answers