Round 833.785 to the nearest hundredth

it should be 1/6 not 16

there are 6 numbers on a die so to get a number you have a 1/6 probability

 so 1/6 x 1/6 = 1/36

the probability of getting both a 6 and a 3 is 1/36

The simplified form of the given expression is [tex]25 x^{2} y^{20}[/tex].

Simplifying expressions mean rewriting the same algebraic expression with no like terms and in a compact manner.

The different Laws of exponents are:

According to the given question.

We have an expression [tex](5xy^{7} )^{2}(y^{2} )^{3}[/tex]

To simplify the above expression we use the exponent rules.

Therefore,

[tex](5xy^{7} )^{2}(y^{2} )^{3}[/tex]

[tex]= 5^{2}x^{2} y^{7\times2} y^{2\times3}[/tex]

[tex]= 25 x^{2} y^{14} y^{6}[/tex]

[tex]= 25 x^{2} y^{14+6}[/tex]

[tex]= 25 x^{2} y^{20}[/tex]

Therefore, the simplified form of the given expression is [tex]25 x^{2} y^{20}[/tex] .

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Answer: 3125

Step-by-step explanation:

Given: A fair sortition trial is carried out, and one of the candidates is assigned the number = 32,041

If each digit can be chosen from 0-4, and if each of the possible sequences is assigned to a candidate, then the number of candidates (repetition of digits allowed)=[tex]5\times5\times5\times5\times5[/tex]

[tex]=3125[/tex]

Hence, the number of candidates there are = 3125.

$4100*(1+0.04/12)^(12*10)

4100*1.490832682=

$6112.41

Answer: Our required probability is 0.34.

Step-by-step explanation:

Since we have given that

Number of seventh grade student s= 21

Number of eight grade students = 14

Number of ninth grade students = 7

Total number of boys and girls = 42

Probability of an eigthth grader winning the competition would be

[tex]\dfrac{14}{42}\\\\=\dfra{1}{3}\\\\=0.3333333..................\\\\\approx 0.34[/tex]

Hence, our required probability is 0.34.

To subtract the mixed numbers 9 1/4 and 6 2/3, first convert them to improper fractions, find a common denominator, and then perform the subtraction. The result is expressed as the mixed number 2 7/12 with the fractional part in lowest terms.

Subtracting Mixed Numbers

The question involves subtracting mixed numbers, specifically 9 1/4 (nine and one quarter) from 6 2/3 (six and two thirds). First, we need to convert these mixed numbers into improper fractions for easier subtraction.

The answer is 2 7/12.

To subtract mixed numbers, find a common denominator. Subtract the fractional part by subtracting the numerators. Subtract the whole numbers as usual. The answer is 8 7/12.

To subtract mixed numbers, we first need to find a common denominator. In this case, the common denominator is 12. Then, we can subtract the fractions by subtracting the numerators and leaving the denominator the same.

For the whole numbers, we simply subtract them as usual.

So, 9 1/4 - 6 2/3 = 8 7/12.

Answer:

I believe it is The Multiplication Property of Equality

Step-by-step explanation:

Since "x" is being divided by 6, we need to do the opposite!

Answer:

multiplication

Step-by-step explanation:

since x/6 is a fraction and fractions represent division we would havw to do the opposite to isolate the x variable. pLeAsE mArK bRaInLiEsT

Answer:

Step-by-step explanation:

Given dimensions of original rectangle , length(l)=3 cm and width(w)=2.5 cm

We know that after dilation with scale factor (k), the dimension of new figure = k times the original dimensions.

Thus width of new rectangle=[tex]4\times2.5=10\ cm[/tex]

length of new rectangle=[tex]4\times3=12\ cm[/tex]

∴The dimensions of the new rectangle will be 10 cm by 12 cm.

Now, Perimeter of original rectangle=[tex]2(l+w)=2(3+2.5)=2(5.5)=11\ cm[/tex]

Thus, Perimeter of new rectangle=[tex]2(4l+4w)=2(12+10)=2(22)=44\ cm[/tex]

⇒ Perimeter of new rectangle=44 cm=[tex]4\times11\ cm[/tex]

∴The new perimeter will be 4 times the original perimeter.

Now, Area of original rectangle=[tex]lw=3\times2.5=7.5\ cm^2[/tex]

Thus, Area of new rectangle=[tex]4l\times4w=12\times10=120\ cm^2[/tex]

Area of new rectangle=[tex]16lw=16(lw)[/tex]

⇒The new area will be 16 times the original area.

Answer:

[tex]\text{64 must be added to the expression }x^2 + 16x\text{ to make it a perfect-square trinomial}[/tex]

Step-by-step explanation:

Given the expression

[tex]x^2+16x[/tex]

we have to tell which value must be added to the above expression to make it a perfect-square trinomial.

Expression: [tex]x^2+16x[/tex]

By completing the square method,

The square of half of the coefficient of x must be added , we get

[tex]x^2+16x+64[/tex]

[tex]x^2+8x+8x+64[/tex]

[tex]x(x+8)+8(x+8)[/tex]

[tex](x+8)^2[/tex]

which is a perfect square.

[tex]\text{Hence, 64 must be added to the expression }x^2 + 16x\text{ to make it a perfect-square trinomial}[/tex]

Answer:

x = 2.75

Step-by-step explanation:

5x - 2 = x + 9

5x = x + 11

4x = 11

The answer is 22 add

Answer:  The values are

[tex]\cos\theta=-\dfrac{1}{2},~~\textup{and}~~\tan\theta=\sqrt3.[/tex]

Step-by-step explanation:  For an angle [tex]\theta[/tex],

[tex]\sin \theta=-\dfrac{\sqrt3}{2},~\pi<\theta<\dfrac{3\pi}{2}.[/tex]

We are given to find the values of [tex]\cos\theta[/tex] and [tex]\tan \theta[/tex].

Given that

[tex]\pi<\theta<\dfrac{3\pi}{2}\\\\\\\Rightarrow \theta~\textup{lies in Quadrant III}.[/tex]

We will be using the following trigonometric properties:

[tex](i)~\cos \theta=\pm\sqrt{1-\sin^2\theta},\\\\(ii)~\tan\theta=\dfrac{\sin\theta}{\cos\theta}.[/tex]

The calculations are as follows:

We have

[tex]\cos\theta=\pm\sqrt{1-\sin^2\theta}\\\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\left(-\dfrac{\sqrt3}{2}\right)^2}\\\\\\\Rightarrow \cos\theta=\pm\sqrt{1-\left(\dfrac{3}{4}\right)}\\\\\\\Rightarrow \cos\theta=\pm\sqrt{\left(\dfrac{1}{4}\right)}\\\\\\\Rightarrow \cos\theta=\pm\dfrac{1}{2}.[/tex]

Since [tex]\theta[/tex] is in Quadrant III, and the value of cosine is negative in that quadrant, so

[tex]\cos\theta=-\dfrac{1}{2}.[/tex]

Now, we have

[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt3}{2}}{-\frac{1}{2}}=\sqrt3.[/tex]

Thus, the values are

[tex]\cos\theta=-\dfrac{1}{2},~~\textup{and}~~\tan\theta=\sqrt3.[/tex]

Since the given theta lies in third quadrant, then you can use the fact that only tangent and cotangent are positive in third quadrant, rest are negative.

The value of cos and tan for given theta is:

[tex]cos(\theta) = -\dfrac{1}{2}\\\\ tan( \theta) = \sqrt{3}[/tex]

If angle lies between 0 to half of pi, then it is int first quadrant.

If angle lies between half of pi to a pi, then it is in second quadrant.

When the angle lies between [tex]\pi[/tex] and [tex]\dfrac{3\pi}{2}[/tex], then that angle lies in 3rd quadrant.

And when it lies from [tex]\dfrac{3\pi}{2}[/tex] and 0 degrees, then the angle is in fourth quadrant.

Only tangent function and cotangent functions.

In first quadrant, all six trigonometric functions are positive.

In second quadrant, only sin and cosec are positive.
In fourth, only cos and sec are positive.

We can continue as follows:

[tex]sin(\theta) = -\dfrac{\sqrt{3}}{2}\\ sin(\theta) = sin(\pi + 60^\circ)\\ \theta = \pi + 60^\circ[/tex]

Thus, evaluating cos and tan at obtained theta:

[tex]cos(\pi + 60^\circ) = -cos(60) = -\dfrac{1}{2}\\ tan( \pi + 60^\circ) = tan(60) = \sqrt{3}[/tex]

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Answer:

Step-by-step explanation:

cot(x) can be written as

[tex]cot(x) =\frac{cosx}{sinx}[/tex]

Here we have [tex]cot(\frac{2x}{3}[/tex]

so It will be undefined whenever

[tex]sin(\frac{2x}{3}) = 0[/tex]

As we cannot have a 0 in the denominator .

so to point all the discontinuties we need to identify the x values where

[tex]sin (\frac{2x}{3} ) = 0[/tex]

[tex]\frac{2x}{3} = 0 + \pi k[/tex] where k is any integer

[tex]x= \frac{3\pi }{2} k[/tex]

It means f(x) is discontinuous at all the values of [tex]x = \frac{3\pi }{2}k[/tex]

for k = 0 , 1 ,2.....

Final answer:

The goal of solving equations with variables on both sides or multi-step equations is to isolate and solve for the variable. Operations such as addition, subtraction, multiplication, and division are used to maintain equation balance and eliminate terms, which is necessary for solving any algebraic equation. The process involves careful step-by-step simplification, including dealing with multiple variables when necessary.

Explanation:

The goal of solving equations with the variable on each side is essentially the same as solving multi-step equations because in both scenarios, we aim to isolate the variable to find its value. When we encounter an equation with variables on both sides, we usually perform operations to get all variable terms on one side and constant terms on the other. This often involves addition or subtraction to eliminate terms and then the use of multiplication or division to isolate the variable. This process aligns with the steps taken in solving multi-step equations, where we also combine like terms, use inverse operations, and apply the distributive property if necessary.

Multiplication or division by the same number on both sides of an equation maintains equality, which is a fundamental principle in algebra. For an equation with more than one term on either side, it is important to enclose the side with more terms in brackets before applying a multiplication or division operation to ensure the operation is distributed to each term within the brackets. By doing so, we preserve the balance of the equation while simplifying it step by step toward finding the unknown value.

When faced with an equation containing more than one variable, our objective is to use algebra to determine the remaining unknowns. This could involve solving simultaneous equations or manipulating a single equation to isolate the variable of interest. In more complex scenarios involving several unknowns, finding a set of equations that each contain only one unknown becomes essential, and careful consideration of physical principles or context may guide us in selecting the right equations to solve for the variables in question.

Answer:

A ⊆ B

Step-by-step explanation:

Answer:

the answer on edge is D) h(3)-h (0)/3

And, the answer to the equation is 156.