Solve. x^-9 y^3 / x^-7 y^8

1 answer:

3 0

$\dfrac{x^{-9}y^3}{x^{-7}y^8}=x^{-9-(-7)}y^{3-8}=x^{-2}y^{-5}=\dfrac{1}{x^2y^5}\\\\Used:\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\a^{-n}=\dfrac{1}{a^n}$

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Triangle A B C is shown. The length of B C is 6 and the length of A B is 2 StartRoot 2 EndRoot. Angle A B C is 80 degrees. Trigo

lawyer [7]

Answer:

8.4 square units

Step-by-step explanation:

See attachment for the figure.

Trigonometric area formula is Area = $\frac{1}{2}\times absin(C)$

Where a and b are two sides of the triangle and C is the angle between these two sides.

Given:

side a = 6 units, c = 2√2 units and ∠B = 80°

Then area of the triangle :

Area = $\frac{1}{2}\times acsin(B)$

= $\frac{1}{2}\times acsin(B)$

= $\frac{1}{2}\times 6\times 2\sqrt{2}\times sin(80)$

= 6√2×sin80°

= 6×(1.414)×0.9848

= 8.36 square units

≈ 8.4 square units

6 0

3 years ago

Write the equation in slope-intercept form of the line that has a slope of 6 and contains the point (1, 1)

svlad2 [7]

$m=6 , \ \ \ (1, 1) \\ \\To \ find \ our \ equation \ we \ will \ use \ the \ formula: \\ \\ y - y _{1} = m(x - x _{1}), wherem \ is \ the \ slope \ and \ (x _{1}, y _{1}) \ is \ the \ point \\ \\m=6 , \ \ x_{1}=1 , \ \ y_{1} = 1\\\\y - 1 = 6(x-1)\\\\y-1=6x -6 \\\\y=6x-6+1\\\\y=6x-5$