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3 years ago

Graph the systems of equations. {2x+y=8 −x+2y=6 Use the Line tool to graph the lines.

Mathematics

1 answer:

Nitella [24]3 years ago

3 0

Answer:

See attachment

Step-by-step explanation:

We want to graph the system

2x+y=8

-x+2y=6

Let us rewrite the lines in slope intercept form:

y=-2x+8

$y=\frac{1}{2}x+3$

The first line has a slope of -2 and a y-intercept of 8

The second line has a slope of 1/2 and a y-intercept of 3

We can no graph the lines as shown in the attachment.

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leva [86]

Answer:

15.7

Step-by-step explanation:

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2 x pi x r(radius). Radius is half the diameter.

5/2=2.5

2 x 3.14 x 2.5

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3 years ago

if a second function is defined by the formula g(x)=2x-7/3, then what is the value of g(f(1))? show the work that leads to your

nydimaria [60]

YOU NEED THE VALUE OF f(1), OR THERE IS MORE MERIT TO THE QUESTION, BUT HERE IS HOW IT WOULD BE BASED ON THE MERITS GIVEN

$g(x) = 2x - \frac{7}{3} \\ g(f(1)) = 2(f(1)) - \frac{7}{3} \\ g(f(1)) = (2 \times f(1)) - ( \frac{7}{3} )$

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3 years ago

What is linear relationship

disa [49]

Step-by-step explanation:

a linear relationship is simply a straight line relationship.

any equation that can be pot in the form of y = mx + b is a linear relationship.

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2 years ago

Regina purchased 1.75 pounds of turkey breast from her local deli for $5.99 per pound. To the nearest cent, how much did she spe

WITCHER [35]

5.99 per pound * 1.75 pounds purchased = $10.4825. If you round it, she spent $10.48

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2 years ago

Read 2 more answers

How do i solve that question?

yawa3891 [41]

a) The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ .

b) The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ .

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ .

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable ( $y, t$ ) in each side of the identity:

$6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t$ (1)

$6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C$

Where $C$ is the integration constant.

By table of integrals we find the solution for each integral:

$-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C$

If we know that $x = 0$ and $y = 1$ , then the integration constant is $C = -2$ .

The solution of this ordinary differential equation is $y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2 } }$ . $\blacksquare$

b) In this case we need to solve a first order ordinary differential equation of the following form:

$\frac{dy}{dx} + p(x) \cdot y = q(x)$ (2)

Where:

$p(x)$ - Integrating factor
$q(x)$ - Particular function

Hence, the ordinary differential equation is equivalent to this form:

$\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x}$ (3)

The integrating factor for the ordinary differential equation is $-\frac{1}{x}$ . $\blacksquare$

The solution for (2) is presented below:

$y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C$ (4)

Where $C$ is the integration constant.

If we know that $p(x) = -\frac{1}{x}$ and $q(x) = x^{2} + \frac{1}{x}$ , then the solution of the ordinary differential equation is:

$y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C$

$y = x\int {x} \, dx + x\int\, dx + C$

$y = \frac{x^{3}}{2}+x^{2}+C$

If we know that $x = 1$ and $y = -1$ , then the particular solution is:

$y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$

The particular solution of the ordinary differential equation is $y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}$ . $\blacksquare$

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

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2 years ago