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3 years ago

HELP PLEASEEE ASAP!!!

Mathematics

2 answers:

ikadub [295]3 years ago

5 0

Answer:

300

Step-by-step explanation:

now answer me spanish question

Marianna [84]3 years ago

5 0

hmmmmajahwiwnejdjdksjsnxnxjdkss

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jack got the expression 7x+1 and then wrote his answer as 1+7x is his answer an equivalment expression how do you know

Ugo [173]

The equation was just flipped around, it will equal the same no matter what.

I hope this helps!! :)

6 0

2 years ago

How tall is a stack of cube shaped blocks whose volumes are 375 cube inches, 648 cube inches, and 1029 cube inches?

djverab [1.8K]

Answer:

L = 25.959 inches

Step-by-step explanation:

Volume of first cube = 375 inch³

Volume of second cube = 648 inch³

Volume of third cube = 1029 inch³

We need to find the length of the stack of the cube shaped block.

We know that,

The volume of a cube = a³ (a is side of a cube)

$a_1=\sqrt[3]{375} \\\\=7.211\ \text{inches}$

$a_2=\sqrt[3]{648 } \\\\=8.653\ \text{inches}$

$a_3=\sqrt[3]{1029} \\\\=10.095\ \text{inches}$

Hence, the total length of the stack is :

L = 7.211 + 8.653 + 10.095

= 25.959 inches

Hence, this is the required solution.

5 0

3 years ago

Describe the location of the point having the following coordinates.

denis23 [38]

we know that

If the point has an nonzero abscissa (x coordinate) then it can be located in any of the four quadrants

If the point has an negative ordinate (y coordinate) then it can be located in Quadrant III or Quadrant IV

therefore

the answer is the option

Quadrant III or Quadrant IV

7 0

3 years ago

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$

$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

The final value of the converges series for all x.

8 0

4 years ago

Samantha is making some floral arrangements. The table show the prices 1/2 dozen of each type of flower.

adelina 88 [10]

Working:
$5.29 + ((2/3) x 3.59) + ((4/3) x 4.79)
= $14.07 (to 2d.p.)

So, she estimated correctly that she will spend around $14 on the flowers but slightly more than $14, so I don't really know if you should say that she estimated correctly or not

5 0

3 years ago