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2 years ago

Which of the following

Mathematics

1 answer:

SIZIF [17.4K]2 years ago

7 0

Solution

For this case we need to find which function is linear

By definition a linear function is given by:

y= m1 x1+ m2 x2+ ..... +b

And for this case we can see that option b satisfy this

y= b

So then the answer is:

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Luke bought 2 used games at the game store if each used game cost $7.85 and he paid with a twenty dollar bill how much change sh

inn [45]

Answer:

$4.30

Step-by-step explanation:

First, find the cost of both games by multiplying the price by 2:

7.85(2)

= 15.7

Then, to find his change, subtract this from 20:

20 - 15.7

= 4.3

So, Luke should get $4.30 back in change.

7 0

3 years ago

Divide 8 1/8 by 7 1/12 simplify the answer and write as a mixed number

vladimir1956 [14]

The division of 8 1/8 by 7 1/12 is 91/136.

<h3>What is division?</h3>

Division simply has to do with reduction of a number into different parts. On the other hand, a mixed number is the number that's made up of whole number and fraction.

Dividing 8 1/8 by 7 1/12 will go thus:

8 1/8 ÷ 7 1/12

Change to improper fraction

65/8 ÷ 85/7

= 65/8 × 7/85

= 91/136

The division will give a value of 91/136.

Learn more about division on:

brainly.com/question/25289437

#SPJ1

4 0

1 year ago

Integrate the following problem:

vazorg [7]

Answer:

$\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C$

Step-by-step explanation:

The integration by parts formula is: $\displaystyle \int udv = uv - \int vdu$

Let's find u, du, dv, and v for $\displaystyle \int e^-^x \cdot cos2x \ dx$ .

$u=e^-^x$
$du=-e^-^x dx$

$dv=cos2x \ dx$
$v= \frac{sin2x}{2}$

Plug these values into the IBP formula:

$\displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$
$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx$

Now let's evaluate the integral $\displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx$ .

Let's find u, du, dv, and v for this integral:

$u=-e^-^x$
$du=e^-^x dx$
$dv=\frac{sin2x}{2} dx$
$v=\frac{-cos2x}{4}$

Plug these values into the IBP formula:

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx$

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

$\displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx$

Let's substitute this back into the first IBP equation.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Simplify inside the brackets.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]$

Distribute the negative sign into the parentheses.

$\displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx$

Add the like term to the left side.

$\displaystyle \int e^-^x \cdot cos2x \ dx + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$
$\displaystyle \frac{5}{4} \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \frac{e^-^x \cdot cos2x}{4}$