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2 years ago

A unit of volume with dimensions 1 unit × 1 unit × 1 unit

Mathematics

2 answers:

Papessa [141]2 years ago

5 0

V=Length*Height*Depth

Length=1 unit

Height=1 unit

Depth=1 unit

Therefore:

V= (1 unit)*(1 unit)*(1 unit)=1 unit³

matrenka [14]2 years ago

4 0

Do:

$V=1u.1u.1u=1u^3$

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Solving proportion 10\8=n\10

Zanzabum

$\frac { 10 }{ 8 } =\frac { n }{ 10 } \\ \\ 10\cdot \frac { 10 }{ 8 } =\frac { n }{ 10 } \cdot 10\\ \\ \frac { 100 }{ 8 } =n$

$\\ \\ n=\frac { 4\cdot 25 }{ 4\cdot 2 } \\ \\ \therefore \quad n=\frac { 25 }{ 2 }$

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2 years ago

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51 time 37 divided by 1 equals to what really need help failing

gizmo_the_mogwai [7]

Answer:

(51*31)/1 = 1581

Step-by-step explanation:

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2 years ago

You and your friend are painting the walls in your apartment. You estimate

Novay_Z [31]

Answer:

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3 years ago

Lim x-1 x2 - 1/ sin(x-2)

balu736 [363]

Answer:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0$

Explanation:

Assuming the correct expression is to find the following limit:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}$

Evaluate the numerator:

$\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}$

Evaluate the denominator:

Since $\lim_{x \to1}sin(x-2)\neq 0$

$\frac{0}{\lim_{x \to1}sin(x-2)}=0$

4 0

2 years ago

Can someone help me with 10 math questions (:

olga2289 [7]

So volue of a cone=1/3 times (area of base[which is a circle]) times height

area of base=area of circle=pi time radius^2
area=pi times 5^2
area=pi times 25=25pi

1/3 times 25pi times 18
25pi times 18 times 1/3
25pi times 18/3
25pi times 6=150pi

the answer is 150π in^3 or 150π cubic inches or C

( to solve, aprox pi to 3.141592 an multiply 150 by 3.141592=471.239 in^3)

3 0

2 years ago