All categories

3 years ago

What is the point of intersection when the system of equations below is graphed on the coordinate plane? -x+y=4 6x+y= -8

2 answers:

7 0

A) -x+y=4
B) 6x+y= -8
Multiplying "A" by -1
A) x -y = -4 then adding it to B
B) 6x + y = -8
7x =-12
x = -12/7 = -1.7142857143
y = 4 + x = 2.2857142857

Did you make a typo because as you can see, the solution is not one of the 4 choices.

Snezhnost [94]3 years ago

7 0

Answer: It is B (-1,3)

Step-by-step explanation: Used photomath

You might be interested in

one taco supreme and two pan pizzas provide 3850 calories. Two taco supremes and one pan pizza provide 3770 calories . Find the

Sedaia [141]

Answer:

Taco supreme is 1230 calories

Pan pizza is 1310 calories

Step-by-step explanation:

Let's denote taco supreme as T and pan pizza az P. We have:

T+2*P = 3850 and 2*T + P = 3770

If we add the equations we get 3*T+3*P = 7620

3*(T+P) = 7620

T+P = 2540 => we can use T=2540-P

Back to one of the first equations we get 2540-P+2*P=3850

P=3850-2540= 1310

=>T= 2540-1310= 1230

4 0

3 years ago

3 true conjectures that students may make concerning even and odd numbers

Elza [17]

Even numbers are divided by two

Even numbers are easier to cope with in math problems

There are only odd prime numbers

6 0

3 years ago

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$

8 0

2 years ago

Naddik [55]

Answer:

what a confusing question?

Step-by-step explanation:

im guessing 3 feet

8 0

2 years ago

Help with this please..... 3=c/-11 a.33 b.-33 c.-14 d.14

djverab [1.8K]

The answer is b or c

5 0

2 years ago