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3 years ago

Which ordered pair is a solution of the equation?

Mathematics

1 answer:

Bas_tet [7]3 years ago

5 0

Answer:

D) Neither.

Step-by-step explanation:

Check. First, use (3, 15).

When x = 3, y = 15. Plug in the corresponding numbers to the corresponding variables in the equation:

y = 7x - 2

(15) = 7(3) - 2

15 = 21 - 2

15 ≠ 19

Next, check (-1, -10).

When x = -1, y = -10. Plug in the corresponding numbers to the corresponding variables in the equation:

-10 = 7(-1) - 2

-10 = -7 - 2

-10 ≠ -9

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WIll mark brainliest asap

Alborosie

Answer:

If the number is x, we can write the following equation:

2 + 10x = 2

10x = 0 so therefore, x must be 0. There is no other number that satisfies this property.

8 0

3 years ago

The slope of a line is 2, and the y-intercept Is O. What is the equation of the line written in slope-Intercept form?

nikdorinn [45]

Answer:

y=2x

Step-by-step explanation:

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8 0

3 years ago

Keith collected the names and ages of all of his classmates and organized them in the ordered pair (name, age).

erastova [34]

Answer:

It is both a relation and a function.

Step-by-step explanation:

Keith collected the names and ages of all of his classmates and organized them in the ordered pair (name, age).

Here, if we consider the name as the input and age is the output, then each and every different input there is a single output.

Because a single person can not have more than one age.

Therefore, it is both a relation and a function. (Answer)

7 0

4 years ago

Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Finger [1]

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$ .

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$ , we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$ { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition: $y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$

4 0

4 years ago

What is the solution?

borishaifa [10]

Answer:

the solution means the answer to whatever question / statement

4 0

3 years ago