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3 years ago

I need help on number 2

Mathematics

1 answer:

postnew [5]3 years ago

6 0

Answer:

Count the dots

Step-by-step explanation:

and you will get the nunber of kids

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Please help with 1, 2, and 3

Akimi4 [234]

Don’t know 1 but 2 is J or 3 because it’s y=mc+b and m=slope and parallel lines have the same slope and 3 is B or -4/5 because if you use the slope formula ( y2-y1/x2-x1) you get the slope (-4/5) and parallel lines have the same slope.

3 0

3 years ago

(Will give Brainliest) Please Helpp I really need it

Rashid [163]

Answer:

Step-by-step explanation:

2x+5.2 + 3x+ 1.2 +2x+6.2= 180

7x+ 12.6=180

7x 12.6 - 12.6=180 -12.6

7x = 167.4

x = 23.914

should equal 60

3 0

3 years ago

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dolphi86 [110]

A circle, of radius r = radius of the sphere

you can think of it in terms of projection the plan projection of the sphere is a circle.

4 0

4 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.