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3 years ago

Please help this work is due today

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

1. 2

2. -6

3. -1/3

4. -9/4

6. 11/8

5. -15/11

Step-by-step explanation:

y2-y1/x2-x1

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Find the prime factorization of the number 12

Vsevolod [243]

First, you must use a factor tree to find the prime factors of the number.

/ \

6 (2)

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(3) (2)

2 , 2 and 3 are the prime factors that make 12

2 x 2 x 3= 12

2^2 x 3= 12

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3 years ago

How many solutions are there to the system of equations? 4x-5y=5 and -.08x+.10y= 0.10No solution 1 solution 2 solutionsInfinite

dsp73

Infinite solutions

hope this helps

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4 years ago

What is the result when the number 100 is decreased by 10%?

Temka [501]

Answer:

10% of 100 is 90

Step-by-step explanation:

100 · 0.10 = 10

100 - 10 = 90

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2 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.