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3 years ago

What is the graph of the linear inequality y≤−3x+4?

Mathematics

1 answer:

tatiyna3 years ago

5 0

Answer:

It would look like the picture I attached at the bottom.

Step-by-step explanation:

We know that the slope is -3 and the y intercept is (0,4) (plugging in 0 for x will get you that point), and then you can just graph an equation like you normally would, using rise/run to go down 3 units for every one unit you go right, and plugging in easy x values to check your work.

It gets a little tricky because the question then adds the inequality, and we see that y is now less than <em>or equal to </em>the original equation.

Since it is less than, we can shade all the values below the graph.

(Also, you should probably note for future reference that if it was just less than, the shading would look the same while the graph itself would be dotted because the values on the line are nor included in the solution set).

Desmos is a great website to use if you're having trouble graphing in the future :)

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The factorization of 8x3 -125 is (2x-5)(jx2 +kx+25)

lbvjy [14]

$\bf ~\hspace{10em}\textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 8x^3-125~~ \begin{cases} 8=2\cdot 2\cdot 2\\ \qquad 2^3\\ 125=5\cdot 5\cdot 5\\ \qquad 5^3 \end{cases}\implies 2^3x^3-5^3\implies (2x)^3-5^3 \\[2em] [2x-5][(2x)^2+(2x)(5)+5^2]\implies (2x-5)(\stackrel{\stackrel{j}{\downarrow }}{4}x^2+\stackrel{\stackrel{k}{\downarrow }}{10}x+25)$

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3 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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2 years ago

What is the solution to the equation <br> 2x-3/4 ?

amm1812

The solution to this equation is in the picture i’ve put below!

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2 years ago

Let p represent any number in the interval (- 18,22). Rewrite the possible values for p in inequality notation.

Nataliya [291]

Answer:

$-18 < p < 22$

Step-by-step explanation:

We are given the following information in the question.

The open interval: (- 18,22).

Let p belong to the given interval then p represents all the values belonging to the given interval.

Open Interval: An open interval is a interval that does not include the end points.

Thus, we can write:

$-18 < p < 22$

is the required inequality form required.

That is p can take values greater than -18 and smaller than 22.

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3 years ago

If if the branding division were to experience the same percentage increase from year 2 to year 3 as they did from year 1 to yea

Sedaia [141]

Since the basis is from year 1 to year 2, calculate first for the difference of their percentages. That would be:

Difference = year 2 - year 1
Difference = 2.32% - 1.1% = 1.22%

We apply this same value of percentage increase from year 2 to year. Thus, the percentage for year 3 is:

% Year 3 = % Year 2 + percentage increase
% Year 3 = 2.32% + 1.22%
% Year 3 = 3.54%

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2 years ago