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Ganezh [65]
3 years ago
8

An equation is shown below:

Mathematics
1 answer:
gayaneshka [121]3 years ago
8 0
6(2x-11)+15=3x+12 Given
12x-66+15=3x+12 Distribution
12x-51=3x+12 Combine like terms
12x=3x+63 addition
9x=63 subtraction
x=7 division

the value of x that makes the equation true is 7.

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The temperature at 12 noon was 10 degree celsius. If it decreases at the rate of 2 degree celsius per hour until midnight, at wh
xenn [34]

9 o clock for 8 degree below zero 14 below zero at midnight

8 0
3 years ago
Need some help with these problems please.
goldenfox [79]

Answer:

See below.

Step-by-step explanation:

a.  

Divide the leading terms:

6x^4 / 2x^2 = 3x^2

3x2 is a parabola so the long run is

as x ---> +/- infinity r(x) ----> + infinity. (answer).

b.

10,000x^3 / 50 x^3

= 200.

So r(x) as a horizontal asymptote at y ( r(x)) = 200.  (answer).

3 0
3 years ago
Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0  } \atop {0}; Otherwise} \right.

The value of constant C can be obtained as:

\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1

\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz  }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ])  } \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}]  } \, dy  }) \, dx =1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0  }) \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}]   } \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}]  } \, dx = 1

50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1

50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1

50C[0+\frac{1}{0.5} ] =1

100C = 1 ⇒ C = \frac{1}{100}

C = 0.01

3 0
3 years ago
No I need 54 inches of wood to build a picture frame. if the wood cost $2.50 a foot, how much will be pay?
tamaranim1 [39]

Answer:

135

Step-by-step explanation:

2 x 54 = 108 and .50 x 54 = 27 so 108 + 27 = 135

7 0
3 years ago
7x+1=8x+3<br><br> Please help I did it but it said my answer was wrong
sweet-ann [11.9K]

I did some math and i got this

x = -2

8 0
2 years ago
Read 2 more answers
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