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3 years ago

How many sections will a paper have if you fold it seven times??? Asking this because my math book says so... :/

1 answer:

7 0

A4 80g copy paper 297mm long ang 0.1mm thick so after 7 folds you would have 2.5mm of length 12.8mm of thickness.

Actually i cant fold it more than 6 times because the thickness going arund in each fold would consume to much.

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Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
$(xy)^a = x^ay^a$

$x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4$

The solution of the expression is $\dfrac{4x^4}{y^6}$ .

<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$ ,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$ .

Learn more about Exponentiation:

brainly.com/question/2193820

8 0

2 years ago

I Need Help With 9 And 10Please Help

Burka [1]

9: the answer is 4 to 13
10: the answer is 6 to 13

5 0

3 years ago

1. Identify your variables using let statements.

Advocard [28]

The number of chicken dish is 131 dishes while the number of lobster dishes is 119 dishes

<h3>How to find the number of each type of dish.</h3>

Let

x = number of chicken dish and
y = number of lobster dish

<h3>How to determine the equations for the linear system</h3>

Since it cost $40 per dish for chicken dish, the total cost of chicken dish is 40x.

Also, since it cost $70 per dish for lobster dish, the total cost of lobster dish is 70y

So, the total cost of dish is T = 40x + 70y

Also, we know that the total cost of dish is $13,570.

So, 40x + 70y = 13570 (1)

Also, there are going to be 250 guest at the dinner and thus 250 dishes.

So, the total number of dishes is x + y = 250

So, x + y = 250 (2)

So, the equations of the linear system are

40x + 70y = 13570 (1)

x + y = 250 (2)

<h3>How to determine the solution for the linear system</h3>

Since

40x + 70y = 13570 (1)

x + y = 250 (2)

From (2), x = 250 - y

Substituting x into (1), we have

40x + 70y = 13570 (1)

40(250 - y) + 70y = 13570 (1)

10000 - 40y + 70y = 13570

30y = 13570 - 10000

30y = 3570

y = 3570/30

y = 119

Substituting y = 119 into x, we have

x = 250 - y

x = 250 - 119

x = 131

Therefore, the number of chicken dish is 131 dishes while the number of lobster dishes is 119 dishes

Learn more about equation of linear system here:

brainly.com/question/13729904

#SPJ1

7 0

2 years ago

1/3 (x + 4 ) = 20 wt the answer

sashaice [31]

Answer:

x = 56

Step-by-step explanation:

To start, distribute the 1/3 onto the x and the 4 in the parenthesis. That leaves you with

1/3x + 4/3 = 20

We want the x value by itself so we're going to subtract the 4/3 from the left and the right side, leaving us with

1/3x + 4/3 = 20 - 4/3

1/3x = 18 & 2/3 (since we're dealing with fractions, I'm going to change the 18 & 2/3 into an improper fraction just to make the next step easier.)

1/3x = 56/3

The last step is to divide the 1/3 off of the x, so x is by itself. To divide a fraction by a fraction, just flip the fraction you're dividing with and multiply.

56/3 x 3/1 = 56 (56 x 3 is 168, and 3 x 1 is 3, so when you reduce 168/3, you get 56)

So, 56 is our answer, because after dividing off the 1/3, we're left with x = 56

3 0

2 years ago

Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0

Alik [6]

Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

$(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0$