All categories

3 years ago

{y=-3x+5 {y=-8x+25 i am so confused

Mathematics

1 answer:

GarryVolchara [31]3 years ago

4 0

Answer:

$(x,y)=(4,-7)$

Step-by-step explanation:

$y= -3x+5\\y=-8x+25$

Since both expressions $-3x+5$ and $-8x+25$ are equal to $y$ , set them equal to each other forming an equation in x.

$-3x+5=-8x+25$

Move variable to the left-hand side and change its sign

$-3x+8x+5=25$

Move constant to the right-hand side and change its sign

$-3x+8x=25-5$

Collect like terms

$5x=25-5$

Subtract the numbers

$5x=20$

Divide both sides of the equation by $5$

$x=4$

Substitute the given value of $x$ into the equation $y=-3 x4+5$

$y=-3x4+5$

Multiply the numbers

$y=-12+5$

Calculate the sum

$y=-7$

The possible solution of the system is the ordered pair $(x,y)$

$(x,y) =(4,-7)$

Check if the given ordered pair is the solution of the system of equations

$-7=-3 x 4+5\\-7=-8 x4+25\\$

Simplify the equalities

$-7=-7\\-7=-8 x 4 +25$

Simplify the equality

$-7=-7\\-7=-7$

Since all of the equalities are true, the ordered pair is the solution of the system

$(x,y)=(4,-7)$

PLEASE MARK ME AS BRAINLIEST

You might be interested in

Amber coaches soccer and volleyball. She coaches both sports for a total of 7 hours each day. The soccer practice lasts 1 hour m

hichkok12 [17]

Amber coaches both sports for 7 hours each day

v = volleyball hours
s = soccerball hours

7 hours in total, in which soccerball is 1 hour more than volleyball:
v + s = 7
s = v + 1

plug in v + 1 for s

v + s = 7
v + (v + 1) = 7

simplify

v + v + 1 = 7
2v + 1 = 7
2v + 1 (-1) = 7 (-1)
2v = 6
2v/2 = 6/2
v = 6/2
v = 3

Plug in 3 for v for one of the equations.

s = v + 1
s = (3) + 1
s = 4

Amber coaches 3 hours for volleyball, and 4 hours for soccer, making a total of 7 hours

hope this helps

6 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 integration by parts:

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

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

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

How many terms are to be considered in the series with first term - 3 and common ratio r = -4 for the sum to exceed 1507

Artemon [7]

Step-by-step explanation:

However you may not have seen the above mentioned in this paragraph world that as long you are can make the most right of you of https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?id=OGC.395456fabbd40f62619c2b35cdee2905&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fbmYTQaG6gk4jS%2f200.gif&ehk=lsVHE7Ou9TrzWjRTYHyB0EbhZaNbCPHhLjUZQhprHK0%3d https://tse1.mm.bing.net/th?i%2fmedia%2fDpB9NBjny7jF1pd0yt2%2fgiphy.gif&ehk=l0INVIjctOcHVK2Ut4YwFkpwLZee5S1ywsPZyrfz8IA%3d https://tse2.mm.bing.net/th?id=OGC.65ac358bb15fab2c01c3ab4bf6ca4362&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fDpB9NBjny7jF1pd0yt2%2fgiphy.gif&ehk=l0INVIjctOcHVK2Ut4YwFkpwLZee5S1ywsPZyrfz8IA%3d https://tse2.mm.bing.net/th?id=OGC.65ac358bb15fab2c01c3ab4bf6ca4362&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fDpB9NBjny7jF1pd0yt2%2fgiphy.gif&ehk=l0INVIjctOcHVK2Ut4YwFkpwLZee5S1ywsPZyrfz8IA%3d https://tse2.mm.bing.net/th?id=OGC.65ac358bb15fab2c01c3ab4bf6ca4362&pid=Api&rurl=https%3a%2f%2fmedia.giphy.com%2fmedia%2fDpB9NBjny7jF1pd0yt2%2fgiphy.gif&ehk=l0INVIjctOcHVK2Ut4YwFkpwLZee5S1ywsPZyrfz8IA%3d https://tse2.mm.bing.net/th?id=OGC.65