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4 years ago

Help, I need to pass this class

Mathematics

1 answer:

Oksana_A [137]4 years ago

5 0

45/9 is 5 it means 1 part is =5
and 5 * 5 is = 25
32 - 25 = 7

its b

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2(x+z) - y x=6 y=8 z=3 need help ASAP PLS

Minchanka [31]

Answer:

Step-by-step explanation:

substitute the given values into the expression

2(x + z) - y

= 2(6 + 3) - 8

= 2(9) - 8

= 18 - 8

= 10

7 0

2 years ago

A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the

Vedmedyk [2.9K]

(a) Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T - 75)$ where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get $\dfrac{dT}{T - 75} = k dt$ . Integrate both sides to get $\ln|T - 75| = kt + C$ .

Since $T(0) = 185$ , we solve for C:
$|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110$
So we get $\ln|T - 75| = kt + \ln 110$ . Use T(30) = 150 to solve for k:
$\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)$

So

$\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t} \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}$

But choose Positive because T > 75. Temp of turkey can't go under.

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)} = 136.929 \approx 137{}^{\circ}F$

(b)

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30} \\
100 = 75 + 110(15/22)^{t/30} \\
25 = 110(15/22)^{t/30} 
\frac{25}{110} = (15/22)^{t/30} \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}$

Dogs of the AMS.

4 0

3 years ago

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

4 years ago

A positive integer from one to six is to be chosen by casting a die. Thus the elements c of the sample space C are 1, 2, 3, 4, 5

Alex_Xolod [135]

Answer with Step-by-step explanation:

We are given that six integers 1,2,3,4,5 and 6.

We are given that sample space

C={1,2,3,4,5,6}

Probability of each element= $\frac{1}{6}$

We have to find that $P(C_1),P(C_2),P(C_1\cap C_2) \;and\; P(C_1\cup C_2)$

Total number of elements=6

$C_1$ ={1,2,3,4}

Number of elements in $C_1$ =4

$P(E)=\frac{number\;of\;favorable \;cases}{Total;number \;of\;cases}$

Using the formula

$P(C_1)=\frac{4}{6}=\frac{2}{3}$

$C_2$ ={3,4,5,6}

Number of elements in $C_2$ =4

$P(C_2)=\frac{4}{6}=\frac{2}{3}$

$C_1\cap C_2$ ={3,4}

Number of elements in $(C_1\cap C_2)=2$

$P(C_1\cap C_2)=\frac{2}{6}=\frac{1}{3}$

$C_1\cup C_2=$ {1,2,3,4,5,6}

$P(C_1\cup C_2)=\frac{6}{6}=1$

4 0

3 years ago

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second