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2 years ago

5

this figure shows a right prism with a trapezoidal base, what is the sum of the areas of the four lateral (rectangular) faces of

the prism in cubic centimeters unit?

1 answer:

kirill [66]2 years ago

4 0

Step-by-step explanation:

the answer is 54 cm

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In an arithmetic sequence, the nth term an is given by the formula an=a1+(n−1)d, where a1 is the first term and d is the commo

Dmitry_Shevchenko [17]

Answer:

$a_{10} = \frac{10}{65536}$

Step-by-step explanation:

The first step to solving this problem is verifying if this sequence is an arithmetic sequence or a geometric sequence.

This sequence is arithmetic if:

$a_{3} - a_{2} = a_{2} - a_{1}$

We have that:

$a_{3} = 40, a_{2} = 10, a_{3} = \frac{5}{2}$

$a_{3} - a_{2} = a_{2} - a_{1}$

$\frac{5}{2} - 10 = 10 - 40$

$\frac{-15}{2} \neq -30$

This is not an arithmetic sequence.

This sequence is geometric if:

$\frac{a_{3}}{a_{2}} = \frac{a_{2}}{a_{1}}$

$\frac{\frac{5}[2}}{10} = \frac{10}{40}$

$\frac{5}{20} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

This is a geometric sequence, in which:

The first term is 40, so $a_{1} = 40$

The common ratio is $\frac{1}{4}$ , so $r = \frac{1}{4}$ .

We have that:

$a_{n} = a_{1}*r^{n-1}$

The 10th term is $a_{10}$ . So:

$a_{10} = a_{1}*r^{9}$

$a_{10} = 40*(\frac{1}{4})^{9}$

$a_{10} = \frac{40}{262144}$

Simplifying by 4, we have:

$a_{10} = \frac{10}{65536}$

3 0

3 years ago

What is the factor of x²-4x+24

DochEvi [55]

Answer:

Their are no possible factors for this expression.

Step-by-step explanation:

3 0

2 years ago

Find the area of each sector. Use exact answers

BigorU [14]

Answer:

B) 54π in²

Step-by-step explanation:

area of full circle = πr² = 81π in²

area of 4π/3 segment = (81)(4π/3)/2π = 54π in²

6 0

2 years ago

Please help!!!<br> Which expression is equivalent to (3m^-1n^2)^4/(2m^-2n)^3

Elza [17]

$\text{Use:}\\\\(a\cdot b)^n=a^nb^n\\\\(a^n)^m=a^{n\cdot m}\\\\\dfrac{a^n}{a^m}=a^{n-m}$

$\dfrac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}=\dfrac{3^4(m^{-1})^4(n^2)^4}{2^3(m^{-2})^3n^3}=\dfrac{81m^{-4}n^8}{8m^{-6}n^3}\\\\=10.125m^{-4-(-6)}n^{8-3}=10.125m^2n^5$

7 0

2 years ago

Read 2 more answers

Help please? I can choose more than one answer I just don’t understand this stuff

Gwar [14]

$...hay.... \\ a + 8 = - \frac{5}{a} \\ a + \frac{5}{a} = - 8 \\ {a }^{2} + 5 = - 8a \\ {a }^{2} + 8a = - 5 \\ henca \: \: opion \: no.1 \: is \: correct$

7 0

3 years ago