Consider U = {x\x is a positive integer greater than 1}.

Solve application problems using radical equations. Vince wants to make a square patio in his yard. He has enough concrete to pa

dmitriy555 [2]

Answer:

3(sqrt42) ft

Step-by-step explanation:

If l is side length of square patio, then area of patio would be l^2.

l^2 = 378

l = sqrt378 = 3(sqrt42)

3 0

3 years ago

How can you use estimation to justify that the answer $7,875 is reasonable? $1,575 X 5 =p $1,575 X 5 = $7,875. So, p = $7,875.

Alisiya [41]

Answer:

p= 1

Step-by-step explanation:

1,575x5 =p p=1

7 0

3 years ago

Solve 20x = 10 for x. A. x = 1/2 B. x = 1.5 C. x = 2 D. x = 10

alexira [117]

Answer:

A. 1/2

Step-by-step explanation:

20x=10

Divide 20 on both sides of the equation to get x by itself

20x=10

___. __

20. 20

x =1/2

5 0

4 years ago

Read 2 more answers

g The completion times for a job task range from 11.8 minutes to 19.4 minutes and are thought to be uniformly distributed. What

nevsk [136]

Answer: 0.5263

Step-by-step explanation:

Given : The completion times for a job task range from 11.8 minutes to 19.4 minutes .

The density function for uniform distribution function for interval [a,b] :-

$f(x)=\dfrac{1}{b-a}$

Then the density function for the given situation:-

$f(x)=\dfrac{1}{19.4-11.8}=\dfrac{1}{7.6}$

The required interval : $16.8-12.8=4$

Now, the probability that it will require between 12.8 and 16.8 minutes to perform the task will be :-

$\dfrac{4}{7.6}=0.526315789474\approx0.5263$

5 0

3 years ago

The sum of the first n terms of an arithmetic series is n/2(3n-5). If the second and fourth terms of the arithmetic series are t

sergiy2304 [10]

Let a be the first term in the arithmetic sequence. Since it's arithmetic, consecutive terms in the sequence differ by a constant d, so the sequence is

a, a + d, a + 2d, a + 3d, …

with the n-th term, a + (n - 1)d.

The sum of the first n terms of this sequence is given:

$a + (a+d) + (a+2d) + \cdots + (a+(n-1)d) = \dfrac{n(3n-5)}2$

We can simplify the left side as

$\displaystyle \sum_{i=1}^n (a+(i-1)d) = (a-d)\sum_{i=1}^n1 + d\sum_{i=1}^ni = an+\dfrac{dn(n-1)}2$

so that

$an+\dfrac{dn(n-1)}2 = \dfrac{n(3n-5)}2$

or

$a+\dfrac{d(n-1)}2 = \dfrac{3n-5}2$

Let b be the first term in the geometric sequence. Consecutive terms in this sequence are scaled by a fixed factor r, so the sequence is

b, br, br ², br ³, …

with n-th term br ⁿ⁻¹.

The second arithmetic term is equal to the second geometric term, and the fourth arithmetic term is equal to the third geometric term, so

$\begin{cases}a+d = br \\\\ a+3d = br^2\end{cases}$

and it follows that

$\dfrac{br^2}{br} = r = \dfrac{a+3d}{a+d}$

From the earlier result, we then have

$n=7 \implies a+\dfrac{d(7-1)}2 = a+3d = \dfrac{3\cdot7-5}2 = 8$

and

$n=2 \implies a+\dfrac{d(2-1)}2 = a+d = \dfrac{3\cdot2-5}2 = \dfrac12$

so that

$r = \dfrac8{\frac12} = 16$

and since the second arithmetic and geometric terms are both 1/2, this means that

$br=16b=\dfrac12 \implies b = \dfrac1{32}$

The sum of the first 11 terms of the geometric sequence is

S = b + br + br ² + … + br ¹⁰

Multiply both sides by r :

rS = br + br ² + br ³ + … + br ¹¹

Subtract this from S, then solve for S :

S - rS = b - br ¹¹

(1 - r ) S = b (1 - r ¹¹)

S = b (1 - r ¹¹) / (1 - r )

Plug in b = 1/32 and r = 1/2 to get the sum :

$S = \dfrac1{32}\cdot\dfrac{1-\dfrac1{2^{11}}}{1-\dfrac12} = \boxed{\dfrac{2047}{32768}}$

6 0

3 years ago