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Leona [35]
3 years ago
12

Evaluate the expression.

Mathematics
2 answers:
MaRussiya [10]3 years ago
8 0

Answer:

1/125

Step-by-step explanation:

5^-3

We know that a^-b = a/a^b

1/5^3

1/125

Illusion [34]3 years ago
5 0

Answer:

\displaystyle  {5}^{ - 3}  =   \boxed{\frac{1}{ 125} }

Step-by-step explanation:

we would like to evaluate the following expression:

\displaystyle  {5}^{ - 3}

recall that,

\displaystyle  {x}^{ - n}  =  \frac{1}{ {x}^{n} }

let x be 5 thus substitute:

\displaystyle  {5}^{ - 3}  =  \frac{1}{ {5}^{3} }

simplify:

\displaystyle  {5}^{ - 3}  =   \boxed{\frac{1}{ 125} }

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What is the difference between 43.68 - 40.67
mote1985 [20]
It would be 3.01 because 40.00-40.00 is 0 then take away the other numbers 3.68- 0.67 is 3.01
7 0
3 years ago
3x – 9y = -39<br> -3x – 7y = -25 systems of elimination
photoshop1234 [79]

Answer:

y= 4 , x= -1

Step-by-step explanation:

3x-9y= -39

-3x-7y= -25

-16y= -64

y= -64/-16

y= 4

substitute y into any equation

3x -9(4) = -39

3x-36= -39

3x= -39+36

3x= -3

x= -3/3

x= -1

7 0
3 years ago
Hlol can u plz solve b no plz plz.....I will mark as brainliest​
zimovet [89]

Answer:

Step-by-step explanation:

Tan \ B =\frac{4}{3} = \frac{opposite \ side \ of B}{adjacent \ side \ of B}

BC = 15 cm

3x = 15

x = 15/3 = 5

AB =4x = 4*5 = 20 cm

Pythagorean theorem,

AC² = AB² + BC²

       = 20²  + 15 ²

       = 400 + 225

       = 625

AC = √625 = 25 cm

AC = 25 cm

Sin \ A = \frac{opposite  \ side \ of \ angle A }{hypotenuse}\\\\  = \frac{BC}{AB}=\frac{15}{25}=\frac{3}{5}

6 0
3 years ago
How many 18-pound patties can she make from 78of a pound of hamburger? We can 7 patties out of 78pound of hamburgers
myrzilka [38]

Answer: 4.3

Step-by-step explanation:

she can make 4.3, 18 pound patties out of a 78 pound of hamburger

3 0
3 years ago
Find the point P on the graph of the function y=√x closest to the point (9,0)
Sphinxa [80]

Answer:

\displaystyle \frac{17}{2}.

Step-by-step explanation:

Let the x-coordinate of P be t. For P\! to be on the graph of the function y = \sqrt{x}, the y-coordinate of \! P would need to be \sqrt{t}. Therefore, the coordinate of P \! would be \left(t,\, \sqrt{t}\right).

The Euclidean Distance between \left(t,\, \sqrt{t}\right) and (9,\, 0) is:

\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}.

The goal is to find the a t that minimizes this distance. However, \sqrt{t^2 - 17 \, t + 81} is non-negative for all real t\!. Hence, the \! t that minimizes the square of this expression, \left(t^2 - 17 \, t + 81\right), would also minimize \sqrt{t^2 - 17 \, t + 81}\!.

Differentiate \left(t^2 - 17 \, t + 81\right) with respect to t:

\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17.

\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2.

Set the first derivative, (2\, t - 17), to 0 and solve for t:

2\, t - 17 = 0.

\displaystyle t = \frac{17}{2}.

Notice that the second derivative is greater than 0 for this t. Hence, \displaystyle t = \frac{17}{2} would indeed minimize \left(t^2 - 17 \, t + 81\right). This t\! value would also minimize \sqrt{t^2 - 17 \, t + 81}\!, the distance between P \left(t,\, \sqrt{t}\right) and (9,\, 0).

Therefore, the point P would be closest to (9,\, 0) when the x-coordinate of P\! is \displaystyle \frac{17}{2}.

8 0
3 years ago
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