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Anestetic [448]
3 years ago
6

4x+8=6x-14Prove: x= 11

Mathematics
1 answer:
Kazeer [188]3 years ago
6 0
4x+8=6x-14
-4x. -4x
8=2x-14
+14 +14
22=2x
/2. /2
11=x
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Integer operations <br> 11. -60<br> 12. 10<br> 13. -9<br> 14. 14<br> 15. 37
vlada-n [284]

Answer:

11. - 60 =  - 49 \\ 12.10 = 12.1 \\ 13. - 9 = 4 \\ 14.14 =  \frac{707}{50}  \\ 15.37 =  \frac{1537}{100}

3 0
3 years ago
Sea level is considered to be 0 feet. A probe released at sea level dropped at a constant rate for 3.25 minutes, reaching an ele
HACTEHA [7]

d = r * t

d/t = r

-35.75 / 3.25 = r

-11 = r

r = -11 ft / min

d = r*t

d = -11 ft/min * 1 min

d = -11 ft

The probe is 11 feet below sea level after 1 minute

5 0
3 years ago
An alrcraft factory manufactures airplane engines. The unit cost C (the cost in dollars to make each airplane engine) depends on
rewona [7]

C(x) should be ;

C(x)=0.9x² - 306x +36,001

Answer:

$9991

Step-by-step explanation:

Given :

C(x)=0.9x^2 - 306x +36,001

To obtain minimum cost :

Cost is minimum when, C'(x) = 0

C'(x) = 2(0.9x) - 306 = 0

C'(x) = 1.8x - 306 = 0

1.8x - 306 = 0

1.8x = 306

x = 306 / 1.8

x = 170

Hence, put x = 170 in C(x)=0.9x²- 306x +36,001 to obtain the

C(170) = 0.9(170^2) - 306(170) + 36001

C(170) = 26010 - 52020 + 36001

= 9991

Minimum unit cost = 9991

5 0
3 years ago
The table shows three unique functions.
Lady_Fox [76]

The true statements about the functions are:

  • g(x) has the maximum greatest value.
  • h(x) has a range of all negative numbers.

<h3>How to determine the true statements?</h3>

The complete question is in the attached image

From the image, we have:

  • g(x) has the greatest maximum value of 7 1/2
  • All the values of h(x) are negative
  • All the values of f(x) are positive

This means that the true statements about the functions are g(x) has the maximum greatest value and h(x) has a range of all negative numbers.

Read more about domain and range at:

brainly.com/question/17019835

#SPJ1

4 0
2 years ago
Consider the oriented path which is a straight line segment L running from (0,0) to (16, 16 (a) Calculate the line integral of t
aalyn [17]

This question is missing some parts. Here is the complete question.

Consider the oriented path which is a straight line segment L running from (0,0) to (16,16).

(a) Calculate the line inetrgal of the vector field F = (3x-y)i + xj along line L using the parameterization B(t) = (2t,2t), 0 ≤ t ≤ 8.

Enter an exact answer.

\int\limits_L {F} .\, dr =

(b) Consider the line integral of the vector field F = (3x-y)i + xj along L using the parameterization C(t) = (\frac{t^{2}-256}{48} ,\frac{t^{2}-256}{48} ), 16 ≤ t ≤ 32.

The line integral calculated in (a) is ____________ the line integral of the parameterization given in (b).

Answer: (a) \int\limits_L {F} .\, dr = 384

              (b) the same as

Step-by-step explanation: <u>Line</u> <u>Integral</u> is the integral of a function along a curve. It has many applications in Engineering and Physics.

It is calculated as the following:

\int\limits_C {F}. \, dr = \int\limits^a_b {F(r(t)) . r'(t)} \, dt

in which (.) is the dot product and r(t) is the given line.

In this question:

(a) F = (3x-y)i + xj

r(t) = B(t) = (2t,2t)

interval [0,8] are the limits of the integral

To calculate the line integral, first substitute the values of x and y for 2t and 2t, respectively or

F(B(t)) = 3(2t)-2ti + 2tj

F(B(t)) = 4ti + 2tj

Second, first derivative of B(t):

B'(t) = (2,2)

Then, dot product between F(B(t)) and B'(t):

F(B(t))·B'(t) = 4t(2) + 8t(2)

F(B(t))·B'(t) = 12t

Now, line integral will be:

\int\limits_C {F}. \, dr = \int\limits^8_0 {12t} \, dt

\int\limits_L {F}. \, dr = 6t^{2}

\int\limits_L {F.} \, dr = 6(8)^{2} - 0

\int\limits_L {F}. \, dr = 384

<u>Line integral for the conditions in (a) is 384</u>

<u />

(b) same function but parameterization is C(t) = (\frac{t^{2}-256}{48}, \frac{t^{2}-256}{48} ):

F(C(t)) = \frac{t^{2}-256}{16}-\frac{t^{2}-256}{48}i+ \frac{t^{2}-256}{48}j

F(C(t)) = \frac{2t^{2}-512}{48}i+ \frac{t^{2}-256}{48} j

C'(t) = (\frac{t}{24}, \frac{t}{24} )

\int\limits_L {F}. \, dr = \int\limits {(\frac{t}{24})(\frac{2t^{2}-512}{48})+ (\frac{t}{24} )(\frac{t^{2}-256}{48})  } \, dt

\int\limits_L {F} .\, dr = \int\limits^a_b {\frac{t^{3}}{384}- \frac{768t}{1152} } \, dt

\int\limits_L {F}. \, dr = \frac{t^{4}}{1536} - \frac{768t^{2}}{2304}

Limits are 16 and 32, so line integral will be:

\int\limits_L {F} \, dr = 384

<u>With the same function but different parameterization, line integral is the same.</u>

6 0
3 years ago
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