All categories

3 years ago

How are a common denominator and a common multiple alike and diferent

Mathematics

2 answers:

marin [14]3 years ago

6 0

Answer:

To find a common denominator of two fractions, you have to find a number that both denominators of the given fractions will divide into. And to find a common multiple, you have to find a number that both given numbers will divide into.

Step-by-step explanation:

docker41 [41]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Write the standard equation for the circle.

Kay [80]

The equation of the circle with center (h, k) and radius r is
.. (x -h)^2 +(y -k)^2 = r^2

1. selection D matches the specification of the circle

2. selection A matches the given equation

3. None of the offered selections matches
.. (x -3)^2 +(y -5)^2 = 74

4. selection A matches the requirements

3 0

3 years ago

Which of the following polygons has five lines of symmetry?

Ivan

Pentagons are and example.

6 0

3 years ago

686.924 to the nearest ten

xenn [34]

<span>686.924 to the nearest ten is 690.0 because the ones place is a number of 5 and above, which means you must move the tens place up 1 value.</span>

6 0

3 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

2 years ago

Solve the equation 3^(2x) -4*3^(x+1) + 27 = 0. use rules of exponents to change 3^(x+1)

Leviafan [203]

Answer:

$\huge\boxed{x=1\ \vee\ x=2}$

Step-by-step explanation:

$3^{2x}-4\cdot3^{x+1}+27=0\\\\\text{use}\ (a^n)^m=a^{nm}\ \text{and}\ a^n\cdot a^m=a^{n+m}\\\\\left(3^x\right)^2-4\cdot3^x\cdot3^1+27=0\\\\\left(3^x\right)^2-12\cdot3^x+27=0\\\\\text{substitute}\ 3^x=t>0\\\\t^2-12t+27=0\\\\t^2-3t-9t+27=0\\\\t(t-3)-9(t-3)=0\\\\(t-3)(t-9)+0\iff t-3=0\ \vee\ t-9=0\\\\t-3=0\qquad\text{add 3 to both sides}\\\boxed{t=3}\\\\t-9=0\qquad\text{add 9 to both sides}\\\boxed{t=9}$

$\text{We return to substitution:}\\\\3^x=t\\\\3^x=3\ \vee\ 3^x=9\\\\3^x=3^1\ \vee\ 3^x=3^2\\\\\boxed{x=1}\ \vee\ \boxed{x=2}$

4 0

3 years ago