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1 year ago

Tyler simplified the expression x³y9. His procedure is shown below.

Mathematics

1 answer:

V125BC [204]1 year ago

5 0

Answer:

(b) both powers should be in the denominator with positive exponents

Step-by-step explanation:

The applicable rule of exponents is ...

a^-b = 1/a^b

The correct simplification is ...

$x^{-3}y^{-9}=\dfrac{1}{x^3}\cdot\dfrac{1}{y^9}=\boxed{\dfrac{1}{x^3y^9}}$

Both variables should have positive exponents in the denominator.

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2 years ago

If the diameter of a circle is 50 yards ,the radius is

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3 years ago

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The coordinates of the endpoints of ST are (-2,3) and (3, y) respectively supposethe distance between s and t us 13 units.What v

igor_vitrenko [27]

Answer:

The values of y would be -9 and 15

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

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S(-2,3) and T(3,y)

substitute the given values in the formula and solve for y

$13=\sqrt{(y-3)^{2}+(3+2)^{2}}$

$13=\sqrt{(y-3)^{2}+25}$

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$169=(y-3)^{2}+25$

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$(y-3)^{2}=144$

take square root both sides

$y-3=(+/-)12$

$y=3(+/-)12$

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The values of y would be -9 and 15

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3 years ago

Which model does the graph represent?

larisa86 [58]

Answer:

Step-by-step explanation:

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B models an exponentially decreasing function.

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2 years ago

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(a) Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 Correct

Luba_88 [7]

Splitting up the interval of integration into $n$ subintervals gives the partition

$\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]$

Each subinterval has length $\dfrac{1-0}n=\dfrac1n$ . The right endpoints of each subinterval follow the sequence

$r_i=\dfrac in$

with $i=1,2,3,\ldots,n$ . Then the left-endpoint Riemann sum that approximates the definite integral is

$\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n$

and taking the limit as $n\to\infty$ gives the area exactly. We have

$\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}$

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3 years ago