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Allisa [31]
3 years ago
8

Find the volume. Round to the nearest tenth.​

Mathematics
1 answer:
sergey [27]3 years ago
3 0

Answer:

a

Step-by-step explanation:

volume of cuboid = 8* 8 *23

volume of cylinder = pi * 4^2 r* 23

total colume = 2628.11

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Find the oth term of the geometric sequence 7, 14, 28, ...
yaroslaw [1]

Answer:

The nth term of the geometric sequence 7, 14, 28, ... is:

a_n=7\cdot \:2^{n-1}

Step-by-step explanation:

Given the geometric sequence

7, 14, 28, ...

We know that a geometric sequence has a constant ratio 'r' and is defined by

a_n=a_1\cdot r^{n-1}

where a₁ is the first term and r is the common ratio

Computing the ratios of all the adjacent terms

\frac{14}{7}=2,\:\quad \frac{28}{14}=2

The ratio of all the adjacent terms is the same and equal to

r=2

now substituting r = 2 and a₁ = 7 in the nth term

a_n=a_1\cdot r^{n-1}

a_n=7\cdot \:2^{n-1}

Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:

a_n=7\cdot \:2^{n-1}

6 0
2 years ago
What is the area of a circle with the diameter of 8
Mars2501 [29]
Area = πr^2
π4^2 =16π
=50.3
5 0
3 years ago
Read 2 more answers
Could I get some help on these please? Will give points.
Zielflug [23.3K]

Answer:

X=-4, Y=22, idk about the CBD

Step-by-step explanation:

4y+2=90 degrees

subtract 2 on both sides you would get

4y=88 divide 4 with 88 and get 22

solve both X's so 5x+6=3x-2

i would subtract 3x first so i can get

2x+6=-2

then subtract 6 on both sides so you would have

2x=-8 divide -8 with 2 and get -4  

6 0
3 years ago
Whats the roots of the following three problems:
klemol [59]

QUESTION 1

The given function is

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

We simplify to get;

f(x)=\frac{x-3}{x+2}

This function is equal to zero when x-3=0

\Rightarrow x=3

QUESTION 2

The given function is

f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}

f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}

We factor to get;

f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}

f(x)=\frac{5(x-1)}{(2x-3)}

This function equals zero when

5(x-1)=0

x=1

QUESTION 3

The given function is

f(x)=\frac{x^3-8}{x^2-6x+8}

We factor to get,

f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}

f(x)=\frac{x^2+2x+4}{x+4}

The function equals zero when x^2+2x+4=0

D=b^2-4ac

D=2^2-4(1)(4)

D=-12

Hence the equation has no real roots.

The complex roots are

x=-\sqrt{3}i-1 or x=-1+\sqrt{3}i

4 0
3 years ago
Please can you help me??
ladessa [460]

Answer:

Why

Step-by-step explanation:

Did you report my answer ?

Did u notice u lost 40 points so then u tricked me into answering it for free ?

4 0
2 years ago
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