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2 years ago

Use absolute value to express the distance between −10 and 16 on the number line.

Mathematics

2 answers:

pashok25 [27]2 years ago

6 0

Answer:

-6 because the rule is that if both numbers are negative or positive it is a positive equation but, here the numbers are only a negative and a positive which means they re a negative.

Step-by-step explanation:

Ede4ka [16]2 years ago

4 0

I think 12,13,14 that's what I think

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Astrid is in charge of building a new fleet of ships. Each ship requires 404040 tons of wood, and accommodates 300300300 sailors

prisoha [69]

Answer:

10 Ships

Step-by-step explanation:

Source: Khan Academy

4 0

2 years ago

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Which represents a quadratic function? f(x) = −8x3 − 16x2 − 4x f (x) = three-quarters x 2 + 2x − 5 f(x) = StartFraction 4 Over x

tatyana61 [14]

Answer:

$2.\ f(x) = \frac{3}{4}x^2 + 2x - 5$

Step-by-step explanation:

Given

$f(x) = -8x^3 - 16x^2 - 4x\\f(x) = \frac{3}{4}x^2 + 2x - 5\\f(x) = \frac{4}{x^2} - \frac{2}{x} + 1\\f(x) = 0x^2 - 9x + 7$

Required

Which of the above is a quadratic function

A quadratic function has the following form;

$ax^2 +bx + c = 0 \ where \ a\neq 0$

So, to get a quadratic function from the list of given options, we simply perform a comparative test of each function with the form of a quadratic function

$1.\ f(x) = -8x^3 - 16x^2 - 4x$

This is not a quadratic function because it follows the form $f(x) = ax^3 + bx^2 + c$ and this is different from $ax^2 +bx + c = 0 \ where \ a\neq 0$

$2.\ f(x) = \frac{3}{4}x^2 + 2x - 5$

This function has an exact match with $ax^2 +bx + c = 0 \ where \ a\neq 0$

By comparison; $a = \frac{3}{4}\ b = 2\ and\ c = -5$

$3.\ f(x) = \frac{4}{x^2} - \frac{2}{x} + 1$

This is not a quadratic function because it follows the form $f(x) = \frac{a}{x^2} + \frac{b}{x} + c$ and this is different from $ax^2 +bx + c = 0 \ where \ a\neq 0$

$4.\ f(x) = 0x^2 - 9x + 7$

This is not a quadratic function because it follows the form $f(x) = ax^2 +bx + c = 0\ but\ a = 0$

Unlike the quadratic function where $a\neq 0$

So, from the list of given options, only $2.\ f(x) = \frac{3}{4}x^2 + 2x - 5$ satisfies the given condition

5 0

2 years ago

A graph of a quadratic function y = f(x) is shown below.

spin [16.1K]

We are given a graph of a quadratic function y = f(x) .

We need to find the solution set of the given graph of a quadratic function .

<em>Note: Solution of a function the values of x-coordinates, where graph cut the x-axis.</em>

For the shown graph, we can see that parabola in the graph doesn't cut the x-axis at any point.

It cuts only y-axis.

Because solution of a graph is only the values of x-coordinates, where graph cut the x-axis. Therefore, there would not by any solution of the quadratic function y = f(x).

<h3>So, the correct option is 2nd option :∅.</h3>

8 0

2 years ago

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What is the solution to the system of equations? y = A system of equations. Y equals StartFraction one-third EndFraction minus 1

lara31 [8.8K]

Answer:

$(6,-8)$ .

Step-by-step explanation:

We have been given a system of equations. We are asked to solve the given system.

$y=\frac{1}{3}x-10...(1)$

$2x+y=4...(2)$

From equation (2), we will get:

$y=4-2x...(2)$

Upon substituting this value in equation (1), we will get:

$4-2x=\frac{1}{3}x-10$

Upon multiply by 3 on both side, we will get:

$3\cdot 4-3\cdot 2x=3\cdot \frac{1}{3}x-3\cdot 10$

$12-6x=x-30$

$12-6x-x=x-x-30$

$12-7x=-30$

$12-12-7x=-30-12$

$-7x=-42$

$\frac{-7x}{-7}=\frac{-42}{-7}$

$x=6$

Upon substituting this in equation (2), we will get:

$y=4-2(6)$

$y=4-12$

$y=-8$

Therefore, the solution for our given system of equations is $(6,-8)$ .

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2 years ago

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H(x)=5x-1h(x)=5x−1, find h(1)h(1

lisabon 2012 [21]

Answer:

Solve f(x)=6x-12 g(x)=5x+1 h(x)=x^2-4 (f+g)(x). f(x)=6x−12 f ( x ) = 6 x - 12 g(x)=5x+1 g ( x ) = 5 x + 1 h(x)=x2−4 h ( x ) = x 2 - 4 (f+g)(x) ( f + g ) ( x ). Evaluate f+g f + ...

Step-by-step explanation:

Solve f(x)=6x-12 g(x)=5x+1 h(x)=x^2-4 (f+g)(x). f(x)=6x−12 f ( x ) = 6 x - 12 g(x)=5x+1 g ( x ) = 5 x + 1 h(x)=x2−4 h ( x ) = x 2 - 4 (f+g)(x) ( f + g ) ( x ). Evaluate f+g f + ...

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2 years ago