I am a quantity that can change or vary, taking on different values. Who am I?

Mathematics

2 answers:

Zigmanuir [339]3 years ago

8 0

The answer is You are a variable

vodka [1.7K]3 years ago

6 0

Obiouly the answer is X because X can change values and could vary anyway is X

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the member increased their elevation 290 feet during their hike this morning. now they are at an elevation of 450 feet

ExtremeBDS [4]

Answer:

160 good luck tho on whatever

Step-by-step explanation:

450-290

3 0

3 years ago

The graph below represents which system of inequalities? graph of two infinite lines that intersect at a point. One line is soli

stiks02 [169]

The first line through points (-3, 0) and (-4, -1) is the orange line.

The second line, through (1, 1) and (2, -1) is the purple line.

Since they are both shaded below, the intersection of the shaded regions, is the region colored in blue.

Both lines are included, since they are solid lines.

The equation of the first line can be found as follows:

the slope is $\frac{0-(-1)}{-3-(-4)}=\frac{1}{1}=1$ , thus, the equation is:
$y-0=1(x-(-4))\\\\y=x+4$

The equation of the second line can be found as follows:

the slope is $\frac{1-(-1)}{1-2}= \frac{2}{-1}=-2$ ,

thus the equation of the second line is :

$y-1=-2(x-1)\\\\y-1=-2x+2\\\\y=-2x+3$

We have to decide on the direction of the inequality sign, so we pick a point, for example (0, 4) which is not in the colored regions of the lines,

so we write the inequalities so that they do not hold for (0, 4)

line 1:

y≥x+4

4≥0+4=4 is true, so we change the sign and write : y≤x+4

similarly, line 2:

y≤-2x+3

4≤-2*0+3=3, which is not true, so we have the correct direction.

Note that since the lines were solid, we include the equality case:

Answer:

the shaded region is the solution to the system of inequalities:

i) y≤x+4
ii) y≤-2x+3

3 0

2 years ago

Read 2 more answers

Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$