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3 years ago

A board that measures 3/4 feet long is cut into 6 equal pieces. What is the length of each piece?

Mathematics

2 answers:

gayaneshka [121]3 years ago

7 0

I would convert from feet to inches and say an 9 in long board is cut.
Now just divide 9 by 6
9/6=1.5 inches

Mice21 [21]3 years ago

6 0

1.5 inches. just do 9 divided by 6

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Divide 4/5 by 2/3. a. 8/15b. 5/6c. 1 1/5d. 1 7/8

Ivan

4/5÷2/3=x
4/5×3/2=x
12/10=x
1 1/5=x

5 0

3 years ago

Please help :) thank you!

const2013 [10]

Answer:

Step-by-step explanation:

just plug in the formula

5 0

3 years ago

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in fourth quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in 4th quadrant so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago

Pls help!!

Allisa [31]

Answer:

Perimeter = 17a

Area = 30 square meters

Step-by-step explanation:

Perimeter = 2b + 8 = 10b + 5 + a = 15ab + 2a - b = 17a

Area = 2 + 8 = 10 + 2 = 12 × 5 = 60 simplify

I honestly don't know if this is correct, I hope it is

8 0

2 years ago

Read 2 more answers

If a positive integer is equal to the following product: 25b3c425b3c4, where b and c are distinct prime numbers greater than 2,

Vlad [161]

Answer: 64 distinct even factors

Step-by-step explanation:

let the 2 distinct numbers be b and c and the integer is expected to be 25b3c425b3c4

since b, c > 2 and prime numbers,

potential options include 3, 5, and 7

hence the likelihoods are (b = 3, c = 5), (b = 5, c = 3), (b = 3, c = 7), (b = 7, c = 3), (b = 5, c = 7), (b = 7, c = 5)

Possibility 1 (b = 3, c = 5)

integer is now 253354253354

the distinct even factors = 2, 202, 262, 1934, 19802, 26462, 195334, 253354, 2000002, 2594062, 19148534, 25588754, 262000262, 1934001934, 2508457954, 253354253354

number of distinct even factors = 16

Possibility 2 (b = 5, c = 3)

integer is now 255334255334

the distinct even factors = 2, 86, 202, 5938, 8686, 19802, 253354, 599738, 851486, 2000002, 25788734, 58792138, 25588754, 86000086, 2528061934, 1934001934, 5938005938, 253354253354

number of distinct even factors = 18

Possibility 3 (b = 3, c = 7)

integer is now 253374253374

the distinct even factors = 2, 6, 22, 66, 202, 242, 606, 698, 726, 2094, 2222, 6666, 7678, 19802, 23034, 24442, 59406, 70498, 73326, 84458, 211494, 217822, 253374, 653466, 775478, 2000002, 2326434, 2396042, 6000006, 6910898, 7188126, 8530258, 20732694, 22000022, 25590774, 66000066, 76019878, 228059634, 242000242, 698000698, 726000726, 836218658, 2094002094, 2508655974, 7678007678, 23034023034, 84458084458,253354253354

number of distinct even factors = 48

Possibility 4 (b = 7, c = 3)

integer is now 257334257334

the distinct even factors = 2, 6, 14, 22, 42, 66, 154, 202, 462, 606, 1114, 1414, 2222, 3342, 4242, 6666, 7798, 12254, 15554, 19802, 23394, 36762, 46662, 59406, 85778, 112514, 138614, 217822, 257334, 337542, 415842, 653466, 787598, 1237654, 1524754, 2000002, 2362794, 3712962, 4574262, 6000006, 8663578, 11029714, 14000014, 22000022, 25990734, 33089142, 42000042, 66000066, 77207998, 121326854, 154000154, 231623994, 363980562, 462000462, 849287978, 1114001114, 2547863934, 3342003342, 7798007798, 12254012254, 23394023394, 36762036762, 85778085778, 257334257334

number of distinct even factors = 64

Possibility 5 (b = 5, c = 7)

integer is now 255374255374

the distinct even factors = 2, 14, 34, 58, 74, 202, 238, 406, 518, 986, 1258, 1414, 2146, 3434, 5858, 6902, 7474, 8806, 15022, 19802, 24038, 36482, 41006, 52318, 99586, 127058, 138614, 216746, 255374, 336634, 574258, 697102, 732674, 889406, 1517222, 2000002, 2356438, 3684682, 4019806, 5128718, 9762386, 12455458, 14000014, 21247546, 25792774, 34000034, 58000058, 68336702, 74000074, 87188206, 148732822, 238000238, 361208282, 406000406, 518000518, 986000986, 1258001258, 2146002146, 2528457974, 6902006902, 8806008806, 15022015022, 36482036482, 255374255374

number of distinct even factors = 64

Possibility 6 (b = 7, c = 5)

integer is now 257354257354

the distinct even factors = 2, 202, 19802, 257354, 2000002, 25992754, 2548061954, 257354257354

number of distinct even factors = 8

From the six possibilities, the highest number of likely distinct even factors is 64

7 0

3 years ago