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ZanzabumX [31]
3 years ago
10

Which shows the expression below simplified?

Mathematics
2 answers:
strojnjashka [21]3 years ago
7 0
.00081/ .0009 = .9 OR  D
Leto [7]3 years ago
5 0
0.00081 / 9(10^−4)

=0.00081 / 9(1/10000)

=0.00081 / 9/10000

=0.9

Answer is is D)

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A trapezoid has an area of 60 square inches. The height of the trapezoid is 5 inches. What is the length of the longer base if t
Vinil7 [7]

Answer:

Step-by-step explanation:

Remark

Let the shorter base = x

Let the longer base = 3x

h = 5

Area = 60

Formula

Area = (b1 + b2)*h /2

Solution

60 = (x + 3x)*5 / 2                Multiply both sides by 2

2*60 = (x + 3x)*5                  Combine like terms

120 = 4x *5

120 = 20x                             Divide by 20

120/20 = x

x = 6

Therefore the two bases are

x = 6

3x = 18

5 0
3 years ago
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Find the number that comes after 144five
Over [174]

Answer:

The number that comes after 144five is:

= 200five.

Step-by-step explanation:

Adding 1 to 144 base 5 will result in:

144

+  1

= 200

b) To obtain the next number that comes after 144five, add 1five to 144five.  Since the numbers are in base 5, 1five added to 4five will result in 0 with 1 carried backward.  When 1 is added to the next 4, the result will be 0 with 1 carried backward.  1 added to 1 = 2, all in base 5.  Figures in base 5 cannot exceed 4.  The usual numbers for a base 5 operation are 0, 1, 2, 3, and 4.

7 0
2 years ago
Help me please i can’t figure it out
julsineya [31]

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5 0
3 years ago
The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r
mash [69]

Answer:

Area = -2.3147

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates (2,0)

The area is represented as:

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Where

$r = 1 + \cos \theta$

and

(a,b) = (2,0)

Substitute values for r, a and b in

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta

Expand

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta

By integratin the above, we get:

Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]

Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]

Substitute 0 and 2 for \theta one after the other

Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}

Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  \frac{-sin(2)(cos(2) + 4) - 6}{4}

Get sin(2) and cos(2) in radians

Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}

Area = \frac{-9.2588}{4}

Area = -2.3147

3 0
2 years ago
I need help with math algebra
Nutka1998 [239]
What is the question you need help with?
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3 years ago
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