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OLga [1]
3 years ago
14

Evaluate the expression,75m+11n, when m=3 and n= 8.

Mathematics
1 answer:
yulyashka [42]3 years ago
5 0

use the substitution method

75m+11n

75(3)+11(8)

225+88

= 313

Answer : 313

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A man starts at a point A and walks 18 feet north. He then turns and walks due east at 18 feet per second. If a searchlight plac
34kurt

Answer:

1/10 per sec

Step-by-step explanation:

When he's walked x feet in the eastward direction, the angle Θ that the search light makes has tangent

tanΘ = x/18

Taking the derivative with respect to time

sec²Θ dΘ/dt = 1/18 dx/dt.

He's walking at a rate of 18 ft/sec, so dx/dt = 18.

After 3seconds,

Speed = distance/time

18ft/sec =distance/3secs

x = 18 ft/sec (3 sec)

= 54ft. At this moment

tanΘ = 54/18

= 3

sec²Θ = 1 + tan²Θ

1 + 3² = 1+9

= 10

So at this moment

10 dΘ/dt = (1/18ft) 18 ft/sec = 1

10dΘ/dt = 1

dΘ/dt = 1/10 per sec

8 0
3 years ago
A national trivia tournament starts with 262,144 teams. With each round, there are half as many teams as before. The number of t
nirvana33 [79]

Answer:

D.) The mathematical range is all real numbers greater than 0. The reasonable range is all whole numbers greater than 0 and less than or equal to 262,144.

Step-by-step explanation:

I don't even know, but yeah.

5 0
3 years ago
Read 2 more answers
Could you please help me on question 5
morpeh [17]
Question plz thxs 
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7 0
3 years ago
Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)
Svet_ta [14]
<h2>Answer:</h2>

\boxed{x=1, \y=-1, \ z=2}

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}

Step 4: solve for z, then for y, then for x:

\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}

-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}

By substituting y=-1 \ and \ z=2 into the first equation, we get the x. So:

x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}

6 0
4 years ago
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Find the least common denominator for these
Roman55 [17]

The common denominator is 3a²

6 0
2 years ago
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