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2 years ago

What is the slope of the line through the points (0,-3) and (2,1)?

Mathematics

1 answer:

Ne4ueva [31]2 years ago

6 0

Answer:

alright so you got 2 points! now this is the slope formula! y2-y1/x2-x1

okay let's label your points! (x1 0,y1 -3) (x2 2,y2 1)

let's follow the formula this makes it so we can solve! 1--3/2-0

I understand you may get a decimal but don't use 2.5 use a fraction because teachers seem to like fractions and general decimals can be hard to use (experience lol) 2.5 is 5/2 so this means your slope is 5/2

this formula works every single time and if you memorize it you can do any other problems with 2 points! please memorize y2-y1/x2-x1

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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:

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

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

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

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

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

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

Thus,

3 0

3 years ago

Subtract the following. Write your answer in standard notation . 18-9.2×10^-4

puteri [66]

Answer:1.799908×10^1

Step-by-step explanation:

18-0.00092

=17.99908

In standard form=1.799908×10^1

8 0

3 years ago

Simplify this expression.

Grace [21]

11x2 would be your answer.

6 0

3 years ago

Read 2 more answers

The principal of a large high school wants to learn about student opinion regarding remote learning versus in person learning. E

jasenka [17]

Answer:

Step-by-step explanation:

the students that are in the survey are in it because they like the subject, rather than picking students from both types of groups, the ones that aren't doing remote learning and the ones that are doing remote learning.

6 0

3 years ago

3. A kite is aloft at the end of a 1500 foot string. The string make an angle of 43° with the

sineoko [7]

Answer:

$h= 1500\,*\,sin(43^o)$

Step-by-step explanation:

Notice that there is a right angle triangle formed with sides as follows:

hypotenuse is the actual 1500 ft string. The acute angle 43 degrees is opposite to the segment that represents the height of the kite from the ground. Therefore, the trigonometric ratio that we can use to find that opposite side to the given angle, is the sine function as shown below:

$sin(43^o) = \frac{opposite}{hypotenuse} \\sin(43^o) = \frac{opposite}{1500}\\opposite = 1500\,*\,sin(43^o)\\h= 1500\,*\,sin(43^o)$

8 0

3 years ago