Answer : 4th option
A equation is given to us and we need to find out its graph from the options . The given equation is ,
→ y = -2x + 1
Comparing this equation to the slope intercept form of the line which is y = mx + c , where m is the slope and c is y intercept. We have ,
So the equation of the line will cut y axis at (0,1) and will have a negative slope .
- Since the slope is negative , it will make an obtuse angle in anti clockwise direction with the x axis .
And the graph which satisfies the above conditions is in 4th option .
<u>Hence</u><u> the</u><u> </u><u>correct</u><u> answer</u><u> is</u><u> </u><u>4</u><u>t</u><u>h</u><u> </u><u>option</u><u> </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em>. </em>
17\20 should be the correct answer
Answer:
Step-by-step explanation:
The Pythagorean theorem is sqrt(a^2 + b^ 2) = c, so:
sqrt(22^2 + 8^2) = c
sqrt(548) = c
23.41 = c
Answer:
- b/a
- 16a²b²
- n¹⁰/(16m⁶)
- y⁸/x¹⁰
- m⁷n³n/m
Step-by-step explanation:
These problems make use of three rules of exponents:
![a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}](https://tex.z-dn.net/?f=a%5Eba%5Ec%3Da%5E%7Bb%2Bc%7D%5C%5C%5C%5C%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D%5C%5C%5C%5Ca%5E%7B-b%7D%3D%5Cdfrac%7B1%7D%7Ba%5Eb%7D%20%5Cquad%5Ctext%7Bor%7D%20%5Cquad%20a%5Eb%3D%5Cdfrac%7B1%7D%7Ba%5E%7B-b%7D%7D)
In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)
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1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.
![\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}](https://tex.z-dn.net/?f=%5Cdfrac%7Bb%5E%7B-2%7D%7D%7Bab%5E%7B-3%7D%7D%3D%5Cdfrac%7Bb%5E%7B-2-%28-3%29%7D%7D%7Ba%7D%3D%5Cdfrac%7Bb%7D%7Ba%7D)
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2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.
![\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B1%7D%7B4ab%7D%5Cright%29%5E%7B-2%7D%3D%5Cdfrac%7B1%7D%7B4%5E%7B-2%7Da%5E%7B-2%7Db%5E%7B-2%7D%7D%3D16a%5E2b%5E2)
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3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.
![\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B4mn%7D%7Bm%5E%7B-2%7Dn%5E6%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%284m%5E%7B1-%28-2%29%7Dn%5E%7B1-6%7D%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%284m%5E3n%5E%7B-5%7D%7D%5Cright%29%5E%7B-2%7D%5C%5C%5C%5C%3D4%5E%7B-2%7Dm%5E%7B-6%7Dn%5E%7B10%7D%3D%5Cdfrac%7Bn%5E%7B10%7D%7D%7B16m%5E6%7D)
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4. This works the same way the previous problem does.
![\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7Bx%5E%7B-4%7Dy%7D%7Bx%5E%7B-9%7Dy%5E5%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%28x%5E%7B-4-%28-9%29%7Dy%5E%7B1-5%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%28x%5E%7B5%7Dy%5E%7B-4%7D%5Cright%29%5E%7B-2%7D%5C%5C%5C%5C%3Dx%5E%7B-10%7Dy%5E%7B8%7D%3D%5Cdfrac%7By%5E8%7D%7Bx%5E%7B10%7D%7D)
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5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.
![\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}](https://tex.z-dn.net/?f=%5Cdfrac%7Bm%5E7n%5E3%7D%7Bmn%5E%7B-1%7D%7D%3D%5Cdfrac%7Bm%5E7n%5E3n%7D%7Bm%7D)