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fomenos
3 years ago
5

Solve the inequality for x. 1/2x + 4 < 3x -1 ​

Mathematics
1 answer:
Contact [7]3 years ago
6 0

Answer:

x>2

Step-by-step explanation:

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|4x−1|=7 PLEEEASE HELP!!!
Alisiya [41]

Answer:

x = 2, - 3/2.

Step-by-step explanation:

|4x - 1| = 7

4x - 1 = 7 or  4x - 1 = -7

4x = 8      or     4x =  -6

x = 2 , x = -3/2.

3 0
4 years ago
A gardener uses 76 ft. Of fencing to enclose a rectangular shaped garden. The width of the garden is 2 ft. Shorter than 3 times
-BARSIC- [3]

Answer:

  • 76 = 2(x + (3x-2))
  • 28 ft

Step-by-step explanation:

Since x represents the length in feet, and the width is 2 ft shorter than 3 times the length, the width is 3x-2.

The perimeter is twice the sum of length and width, so we can write the equation ...

  76 = 2(x + (3x-2))

  38 = 4x -2 . . . . . divide by 2, simplify

  40 = 4x . . . . . . . .add 2

  10 = x . . . . . . . . . the length of the garden is 10 ft

The width of the garden is ...

  3·10 -2 = 28 . . . feet

The width of the garden is 28 feet.

5 0
3 years ago
If the length of a rectangle is seven feet less than the width and the perimeter is 79 feet, find the width of the rectangle. ?​
IgorLugansk [536]

Answer:

Step-by-step explanation:2

5 0
4 years ago
Evaluate the following iterated integral by converting to polar coordinates.
saul85 [17]

Looks like the integral is

\displaystyle\int_{-8}^8\int_0^{\sqrt{64-x^2}}\sin(x^2+y^2)\,\mathrm dy\,\mathrm dx

The integration region is the upper half of a circle in the <em>x</em>-<em>y</em> plane centered at the origin with radius 8. So, in polar coordinates, the integral is

\displaystyle\int_0^{2\pi}\int_0^8r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^8r\sin(r^2)\,\mathrm dr

=\pi\displaystyle\int_0^{64}\sin s\,\mathrm ds

where we substituted s=r^2 and \mathrm ds=2r\,\mathrm dr, which gives

=\boxed{\pi(1-\cos(64))}

6 0
3 years ago
Set up the integral that uses the method of disks/washers to find the volume V of the solid obtained by rotating the region boun
Blababa [14]

Answer:

a)\: \frac{3}{2} \pi

b) \:\frac{47}{5} \pi

Step-by-step explanation:

y=3\sqrt x,\:\:y=3,\:\:x=0

a) about the line y = 3

3\sqrt x=3  ⇒  x = 1  is the intersection point

So,

V=\int\limits^1_0\pi(3-3\sqrt x)^2\:dx=\pi\int\limits^1_0(9-18\sqrt x+9x)\:dx=\\\\=\pi(9x-12x^{3/2}+9/2x^2)|^1_0=\pi(9-12+9/2)=\frac{3}{2} \pi

b)  about the line x = 5

y=3\sqrt x  ⇒  x=y^2/9

So,

V = \int\limits^3_0\pi([5-0]^2-[5-y^2/9]^2)\:dy=\pi\int\limits^3_0(25-25+10y^2/9-y^4/81)\:dy=\\\\=\pi(10y^3/27-y^5/405)|^3_0=\pi(10-3/5)=\frac{47}{5} \pi

8 0
4 years ago
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