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3 years ago

1000 go into 3 how many times

Mathematics

2 answers:

sineoko [7]3 years ago

5 0

0 it isn't possible but if you are trying to make it a decimal it would be able to go in 0.0003 times.

Nikolay [14]3 years ago

4 0

0, not possible for 100 to go in to 3

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Find the approximate side length of a square game board with an area of 157 in^2.

Nesterboy [21]

Answer:

12.52 in.

Step-by-step explanation:

Area of square(A)=157in^2

length of square (l)=?

we know,

A=l^2

or,sqrt 157=l

12.52 in=l

3 0

3 years ago

Help plzzz<br>Select the correct answer.<br>

Morgarella [4.7K]

<h3>Answer: C) 3</h3>

=========================================================

Explanation:

f(x) is the outer function, so the final output -8 corresponds to f(x)

We see that f(-4) = -8 in the first column of the table. I'm starting with the output and working my way backward to get the input. So we started with -8 and worked back to -4.

Then we move to the g(x) function to follow the same pattern: start with the output and move to the input. We start at -4 in the g(x) bubble and move to 3 in the x bubble.

In short, g(3) = -4

So,

f(g(x)) = f(g(3)) = f(-4) = -8

We see that x = 3 leads to f(g(x)) = -8

3 0

3 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term $x^2$ , which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression: $x_v=\frac{-b}{2a}$

Since in our case $a=1$ and $b=5$ , we get that the x-position of the vertex is: $x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

$y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}$

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the $x_v=-\frac{5}{2}$ . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.