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3 years ago

How to graph -7x+25y=231

Mathematics

1 answer:

adoni [48]3 years ago

7 0

Answer:

Rewrite the equation as:

y = (231 + 7x)/25

Then plug it into a graphing calculator

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AP STATISTICS

Stels [109]

Based on the random sample of beach lengths taken, the probability of a randomly selected beach having a length of 12 miles is C. 0.

<h3>What is the probability that a beach is 12 miles in length?</h3>

Beach length is considered a continuous variable which takes a numerically positive form because it can have any one of infinite possible values.

As a result, there is no possibility that a beach chosen at random will have a given length of 12 miles which means that the probability is 0.

In conclusion, the probability that a selected beach has a length of 12 miles is 0.

Find out more on continuous variables at brainly.com/question/27761372

#SPJ1

5 0

1 year ago

Beau's recipe fro granola bars calls for 3 1/2 cups of oatmeal. He only has 1/6 scoop. How many scoops of oatmeal will he need t

natulia [17]

$\boxed{21}$ scoops of oatmeal will he need to complete the recipe.

Further explanation:

Given:

Beau's recipe from granola bars calls for $3\dfrac{1}{2}$ cups of oatmeal.

He only is $\dfrac{1}{6}$ scoop.

Explanation:

The cups of oatmeal called for the recipe is $3\dfrac{1}{2}.$

Solve the improper fraction $3\dfrac{1}{2}.$

$\begin{aligned}{\text{Fraction}}&=\frac{{2 \times 3 + 1}}{2}\\&=\frac{{6 + 1}}{2}\\&=\frac{7}{2}\\\end{aligned}$

The Beau has $\dfrac{1}{6}{\text{ scoop}}.$

The number of scoops required to complete the recipe can be obtained as follows,

$\begin{aligned}N&=\frac{{{\text{total cups of oatmeal}}}}{{{\text{the fraction of scoop Beau has}}}}\\&=\frac{{\frac{7}{2}}}{{\frac{1}{6}}}\\&= \frac{7}{2} \times6\\&=\frac{{42}}{2}\\&= 21\\\end{aligned}$

Number of scoops required to complete the oatmeal is $21{\text{ scoops}}.$

Therefore, $\boxed{21}$ scoops of oatmeal will he need to complete the recipe.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Fractions

Keywords: fraction, Beau’s recipe, granola bars, calls, 3 1/2 cups of oatmeal, scoops, recipe, ratio, 1/6 scoop, complete the recipe.

5 0

3 years ago

The graph below shows four sections: A, B, C, and D.

Gemiola [76]

It’s c
Hope this helps

4 0

3 years ago

What is equivalent to (3 - 2x) (4x - 7)

bonufazy [111]

Answer:

-8x^2+26x-21

Step-by-step explanation:

(3-2x)(4x-7)

12x-8x^2-21+14x

-8x^2+12x+14x-21

-8x^2+26x-21

7 0

3 years ago

Read 2 more answers

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

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

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

2 years ago