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2 years ago

Is the origin a solution to the system graphed?

Mathematics

1 answer:

lapo4ka [179]2 years ago

4 0

Answer:

Step-by-step explanation:

I do not believe the origin is a solution.

-> While the blue shaded portion does cover the origin, it is a dashed line meaning that the values on the line are not within the solution to the system graphed

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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Bethany earned a total of $175 for babysitting over her summer break. Which expression shows how much money Bethany earned each

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Answer:

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Step-by-step explanation:

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3 years ago

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Which of the following numbers is closest to the product of 48.9x21.2

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3 years ago

For f(x)=3x+1 and g(x)=x²-6, find (f-g)(x)

Lorico [155]

The value of (f - g)(x) is -x² + 3x + 7.

<h3>What is Function?</h3>

The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

Here, the given functions are;

f(x) = 3x + 1

g(x) = x² - 6

Now,

(f - g)(x) = f(x) - g(x)

= (3x + 1) - (x² - 6)

= 3x + 1 - x² + 6

(f - g)(x) = -x² + 3x + 7

Thus, the value of (f - g)(x) is -x² + 3x + 7.

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brainly.com/question/12431044

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2 years ago

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If the radius of the circle is multiplied by a factor of 3, then the area of the circle is multiplied by a factor of

Oksi-84 [34.3K]

$\bf \textit{area of a circle}\\\\
A=\pi r^2\quad 
\begin{cases}
r=radius\\
-----\\
r=\stackrel{r\times 3}{3r}
\end{cases}\implies A=\pi (3r)^2
\\\\\\
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notice, if the original area was πr², then the new area has a factor in front of it, that much.

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3 years ago

4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25

jarptica [38.1K]

Answer with explanation:

$\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}$

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

$=e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}$

Multiplying both sides by Integrating Factor

$x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)$

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

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