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2 years ago

What does not represent an integer

Mathematics

2 answers:

givi [52]2 years ago

8 0

Fractions and decimals

Veronika [31]2 years ago

8 0

Answer:

An integer is any rational number without a decimal or fraction. This includes negative numbers, and 0. But decimals and fractions are not integers.

Step-by-step explanation:

Brainliest, please!

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When testing for current in a cable with eight color-coded wires, the author used a meter to test five wires at a time. How man

balu736 [363]

Answer:

56 different tests

Step-by-step explanation:

Given:

Number of wires available (n) = 8

Number of wires taken at a time for testing (r) = 5

In order to find the number of different tests required for every possible pairing of five wires, we need to find the combination rather than their permutation as order of wires doesn't disturb the testing.

So, finding the combination of 5 pairs of wires from a total of 8 wires is given as:

$^nC_r=\frac{n!}{r!(n-r)!}$

Plug in the given values and solve. This gives,

$^8C_5=\frac{8!}{5!(8-5)!}\\\\^8C_5=\frac{8\times 7\times 6\times 5!}{5!\times 3\times2\times1}\\\\^8C_5=56$

Therefore, 56 different tests are required for every possible pairing of five wires.

6 0

2 years ago

Solve the system by substitution. 3x + 2y = 15 y = 6x

olga2289 [7]

Answer:

the answer should be (1,6)

8 0

2 years ago

Read 2 more answers

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

1 year ago

A rectangular soccer field is twice as long as it is wide. If the perimeter of a soccer field is 300 yards, what are the fields'

Masteriza [31]

A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards , what are its dimensions?
I know the basic formula is 2W+2L=300 but i am not sure where to go from there...
-----
Equations:
2W + 2L = 300
L = 2W
----
Substitute for "L" and solve for "W":
2W + 2(2W) = 300
6W = 300
W = 50 yds (width)
----
Solve for "L":
L = 2W
L = 100 yds (length)
========================
Cheers.

8 0

2 years ago

Is 3(4x-2) and 12x-2 equivalent?

Korolek [52]

Answer:

Step-by-step explanation:

3(4x - 2) = 12x - 4 ≠ 12x -2

3 0

2 years ago

Read 2 more answers