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Greeley [361]
3 years ago
12

How would the expression x^3-3sqrt(3) be rewritten using difference of cubes?

Mathematics
2 answers:
Semenov [28]3 years ago
3 0
The difference of two cubes is factored by
x^3-y^3=(x-y)(x^2+xy+y^2)

In this case it would be

C) 
iogann1982 [59]3 years ago
3 0

Answer:

Correct option is:

C. (x-\sqrt{3})(x^2+3+x\sqrt{3} )

Step-by-step explanation:

x^3-3sqrt(3)

=  x^3-3\sqrt{3} \\\\=x^3-\sqrt{3}^3

= (x-\sqrt{3})(x^2+3+x\sqrt{3} )

(since, a^3-b^3=(a-b)(a^2+b^2+ab))

Hence, the correct option is:

   C.    (x-\sqrt{3})(x^2+3+x\sqrt{3} )

You might be interested in
400÷6= show how you get 66r6
Aleksandr [31]

Answer:

 0 6 6

   6 ⟌ 4 0 0

     - 0    

       4 0  

     - 3 6  

         4 0

     -   3 6

           4

Step-by-step explanation:

8 0
3 years ago
A magician is performing for an audience of 32 men and 20 women. He selects a volunteer from the audience, has the volunteer sta
g100num [7]
The correct answer is A) 37.4%.

The probability that the first volunteer is a man would be 32/52, since there are 32 men out of 52 people.  The probability that the second volunteer would also be a man would be 31/51, since there is 1 fewer man and 1 fewer person total to choose from.

32/51(31/51) = 992/2652 = 0.374 = 37.4%.
8 0
4 years ago
6c to te second power - 5d + 8
mina [271]
Lets see,
6c to the second power is 36c because 6c times 6c is 36 c
And since the variables are different and only one number doesn't have a variable, 
The answer would be, 36c-5d+8
7 0
4 years ago
Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0
3 years ago
How many solutions does the equation below have.<br> *The answer choices are below*
Natali [406]
<h3>Answer: D) One solution</h3>

==========================================

Work Shown:

2(b+3)-17 = 3b-7+b

2b+6-17 = 4b-7

2b-11 = 4b-7

-11+7 = 4b-2b

-4 = 2b

2b = -4

b = -4/2

b = -2

There is one solution and it is b = -2

A quick way to tell we have one solution is to note that both sides are a linear expression, and that the slopes of each linear expression are different values. If you had two lines with the same slope, then you'd have either parallel lines or coinciding lines (leading to no solutions and infinitely many solutions respectively). When you have two lines with two different slopes, then they are guaranteed to intersect only once. That intersection point is the solution.

5 0
3 years ago
Read 2 more answers
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