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kipiarov [429]
2 years ago
8

Given the pattern: 18, 9, 4.5, 2.25 Find the next two terms

Mathematics
1 answer:
SIZIF [17.4K]2 years ago
8 0

Answer:

1.125 and 0.5625

Step-by-step explanation:

Hello there!

It seems that every term is dividing by 2 to get the next term

18/9=2

9/4.5=2

So to find the next terms we just divide the previous term by 2

2.25/2=1.125

1.125/2=0.5625

so the next two terms would be 1.125 and 0.5625

<em>~TheMathWiz</em>

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crosses x-axis at (2, 0 ) and y-axis at (0, - 4 )

To find where the graph crosses the x and y axes ( intercepts )

• let x = 0, in the equation for y- intercept

• let y = 0, in the equation for x- intercept

x = 0 : y = 0 - 4 = - 4 ⇒ (0, - 4 )

y = 0 : 2x - 4 = 0 ⇒ 2x = 4 ⇒ x = 2 ⇒ (2, 0 )


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3 years ago
Help ASAP please
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Step-by-step explanation:

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2 years ago
Sofia jarred 18 liters of jam after 6 days. How many days does Sofia need to spend making jam if she wants to jar 30 liters of j
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Step-by-step explanation:

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A length of rope is stretched between the top edge of a building and a stake in the ground. The rope also touches a tree halfway
alexgriva [62]
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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
2 years ago
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