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3 years ago

What is the volume of a tuna can if the diameter is 13cm and the height is 5cm ?

Mathematics

1 answer:

zloy xaker [14]3 years ago

7 0

The volume of the tuna can is 663.66 cm^3.

Hope that helps!

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3 a plus b equals 54, when b equals 9?

Anton [14]

Answer:

3a+b= 54

3a+9=54

3a=54-9

3a=45

a=45/3

a= 15

8 0

3 years ago

Read 2 more answers

Describe two ways to tell if a graph is a function

Dima020 [189]

Answer:

Inspect the graph to see if any vertical line drawn would intersect the curve more than once. If there is any such line, the graph does not represent a function. If no vertical line can intersect the curve more than once, the graph does represent a function.

Step-by-step explanation:

Hope that helps you out!

I wouldnt mind if you marked brainliest hehehe

Have a good day!

8 0

2 years ago

Let U = {1, 2, 3, 4, 5, 6, 7), A = {2, 4, 6, 7), and B = {1, 4, 7}. Find the set (A U B)'.

den301095 [7]

Step-by-step explanation:

(A U B) = {1,2, 4, 6, 7}

(A U B)'= U- (A U B) = { 3,5}<------- Answer

3 0

3 years ago

(-2,2) lies in what quadrant

Vitek1552 [10]

It lies in quadrant 2 .

4 0

2 years ago

(a) Determine a cubic polynomial with integer coefficients which has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Keith_Richards [23]

Answer:

(a) $x\³ - 6x - 6$

(b) Proved

Step-by-step explanation:

Given

$r = $\sqrt[3]{2} + \sqrt[3]{4}$$ --- the root

Solving (a): The polynomial

A cubic function is represented as:

$f = (a + b)^3$

Expand

$f = a^3 + 3a^2b + 3ab^2 + b^3$

Rewrite as:

$f = a^3 + 3ab(a + b) + b^3$

The root is represented as:

$r=a+b$

By comparison:

$a = $\sqrt[3]{2}$

$b = \sqrt[3]{4}$$

So, we have:

$f = ($\sqrt[3]{2})^3 + 3*$\sqrt[3]{2}*\sqrt[3]{4}$*($\sqrt[3]{2} + \sqrt[3]{4}$) + (\sqrt[3]{4}$)^3$

Expand

$f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

Evaluate like terms

$f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)$

Recall that: $r = $\sqrt[3]{2} + \sqrt[3]{4}$$

So, we have:

$f = 6 + 6r$

Equate to 0

$f - 6 - 6r = 0$

Rewrite as:

$f - 6r - 6 = 0$

Express as a cubic function

$x^3 - 6x - 6 = 0$

Hence, the cubic polynomial is:

$f(x) = x^3 - 6x - 6$

Solving (b): Prove that r is irrational

The constant term of $x^3 - 6x - 6 = 0$ is -6

The divisors of -6 are: -6,-3,-2,-1,1,2,3,6

Calculate f(x) for each of the above values to calculate the remainder when f(x) is divided by any of the above values

$f(-6) = (-6)^3 - 6*-6 - 6 = -186$

$f(-3) = (-3)^3 - 6*-3 - 6 = -15$

$f(-2) = (-2)^3 - 6*-2 - 6 = -2$

$f(-1) = (-1)^3 - 6*-1 - 6 = -1$

$f(1) = (1)^3 - 6*1 - 6 = -11$

$f(2) = (2)^3 - 6*2 - 6 = -10$

$f(3) = (3)^3 - 6*3 - 6 = 3$

$f(6) = (6)^3 - 6*6 - 6 = 174$

For r to be rational;

The divisors of -6 must divide f(x) without remainder

i.e. Any of the above values must equal 0

<em>Since none equals 0, then r is irrational</em>

3 0

2 years ago