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3 years ago

Simply by using rationalize denominators. 2a^-1/2

1 answer:

8 0

Pemdas, so exponent before multiply
so do a^-1/2 then multiply it by 2

remember that $x^ \frac{m}{n}= \sqrt[n]{x^m}$ and
$x^{-m}= \frac{1}{x^m}$
therefor, $a^ \frac{-1}{2}= \frac{1}{a^ \frac{1}{2} } = \frac{1}{ \sqrt{a} }$
the we multiply by 2

$\frac{2}{ \sqrt{a} }$

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What is the solution to the system of equations? 3x + 10y = -47 5x- 7y = 40 Please and thank u

anastassius [24]

Answer:

x=1, y=-5

Step-by-step explanation:

Given equations are:

$3x+10y=-47\ Eqn\ 1\\and\\5x-7y=40\ Eqn\ 2$

In order to solve the equation

Multiplying Eqn 1 by 5 and eqn 2 by 3 and subtracting them

So,

Eqn 1 becomes

15x+50y=-235

Eqn 2 becomes

15x-21y=120

Subtracting 2 from a

15x+50y - (15x-21y) = -235-120

15x+50y-15x + 21y = -355

71y = -355

y = -355/71

y =-5

Putting y= -5 in eqn 1

3x+10(-5) = -47

3x -50 = -47

3x = -47+50

3x = 3

x = 3/3

x = 1

Hence the solution is:

x=1, y=-5

5 0

3 years ago

Read 2 more answers

In a right triangle, the ____________ of an angle can be found by dividing the length of the opposite leg by the length of the t

damaskus [11]

The correct answer for the completion exercise shown above is: sine.

Therefore, the complete text is shown below: "In a right triangle, the sine of an angle can be found by dividing the length of the opposite leg by the length of the triangle's hypotenuse".

A right triangle is a triangle that has an angle of 90 degrees.

The sine is one of the most common trigonometric functions. Therefore, you have that the sine of an angle is:

Sin(α)=opposite/hypotenuse

3 0

3 years ago

A line includes the points (3,4)and (6,10). What is its equation in slope- intercept form

chubhunter [2.5K]

10-4=6. 6-3=3
6÷3 =2. Slope is positive and equals 2

3 0

3 years ago

Urgent. There are 12 coloured counters in a bag. The counters are black, white or grey A counter is chosen at random. The probab

Nastasia [14]

Answer:

1/12

Step-by-step explanation:

Needed information

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

The sum of the probabilities of all outcomes must equal 1

Solution

We are told that the probability that the counter is not black is 3/4.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is black by subtracting 3/4 from 1:

$\sf P(counter\:not\:black)=\dfrac34$

$\implies \sf P(counter\:black)=1-\dfrac34=\dfrac14$

We are told that the probability that the counter is not white is 2/3.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is white by subtracting 2/3 from 1:

$\sf P(counter\:not\:white)=\dfrac23$

$\implies \sf P(counter\:white)=1-\dfrac23=\dfrac13$

We are told that there are black, white and grey counters in the bag. We also know that the sum of the probabilities of all outcomes must equal 1. Therefore, we can work out the probability the counter is grey by subtracting the probability the counter is black and the probability the counter is white from 1:

$\begin{aligned}\sf P(counter\:grey) & = \sf1-P(counter\:black)-P(counter\:white)\\\\ & =\sf 1-\dfrac14-\dfrac13\\\\ & = \sf \dfrac{12}{12}-\dfrac{3}{12}-\dfrac{4}{12}\\\\ & = \sf \dfrac{12-3-4}{12}\\\\ & = \sf \dfrac{5}{12}\end{aligned}$

7 0

2 years ago

A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03% chlorine. Starting at tequals0, city water conta

Katarina [22]

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.

Remember, 1 h = 60 minutes. The initial value problem is

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x = - 6* 10⁻⁴*(3x - 10⁻¹) x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C or |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

7 0

3 years ago