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3 years ago

10

an office supply company sells two types of printers. they charge $95 for one of the printers and $125 for the other. if the com

pany sold 32 printers for a total of$3340 last month, how many of each type were sold

1 answer:

Ivenika [448]3 years ago

4 0

Answer:

22 of the 95$ ones and 10 of the 125

Step-by-step explanation:

22 times 95 = 2090

10 times 125 = 1250

2090+1250=3340

hope this helped

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What is the square root of 5

Zigmanuir [339]

Answer:

2.236

Step-by-step explanation:

you could estimate it to be between 2 and 3 (because 5 is between 4 and 9)

or you just put it into a calculator

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3 years ago

What is the inverse operations of -5x + 10 = 40

noname [10]

Answer:

x=-6

Step-by-step explanation:

−5x+10=40

−5x+10−10=40−10

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−5x/5=30/5

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3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

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3 0

2 years ago

A container that holds sugar is shaped like a cylinder. The radius of the container is 3 inches,

kicyunya [14]

Answer:

B

Step-by-step explanation:

A cylinder is like circles staked on top of each other so you want to find the are of the top circle which is 28.26 because pi r^2 (3^2)*3.14. now you found the area of the top circle. The height is 10.5 so that means there are 10.5 layers so you multiply 28.26 by 10.5 i know its confusing im new to brainly but hopefully this answer helped you You will get 296.73 because i rounded pi to 3.14 but the answer is B

8 0

3 years ago

I need #14 done by tomorrow around 9-10:00. Thank you so much if you do. God bless you :3

Vinvika [58]

Answer:

A) $\frac{1}{5}$

B) - 5

C) Not Possible

D) 5

E) $\frac{-1}{5}$

Step-by-step explanation:

All integers are rational numbers. But not all rational numbers are integers.

All whole numbers are integers. But not all integers are whole numbers.

I am a rational number but not an integer. Located on the right of 0.

This means that it should be a positive number. Since, it is a rational number but not an integer, it should be of the form $\frac{p}{q}$ .

From, the options $\frac{1}{5}$ would fit this description.

I am a rational number and an integer but not a whole number.

This means that it should be a negative integer. Since, all positive integers and zero would be whole numbers. From the options, the answer would be -5.

I am a whole number but not an integer.

This is clearly not possible because all whole numbers are a subset of integers.

I am a rational number, a whole number and an integer.

This means it is a positive integer. 5 would fit this description.

I am a rational number but not an integer; located on the left side of 0.

This means it is a negative number. $-\frac{1}{5}$ should be the answer.

3 0

3 years ago