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3 years ago

Write the equation of a circle with center at (3, -2), which also passes through the point (0, 2).

1 answer:

3 0

$\text{The equation of acircle:}\\\\(x-h)^2+(y-k)^2=r^2\\\\(h;\ k)-center\\r-radius\\\\\text{We have center}\ (3,\ -2)\to h=3\ \text{and}\ k=-2.\\\\\text{The radius is the distance between the points} (3,\ -2)\ \text{and}\ (0,\ 2).\\\\\text{The formula of a distance between two points:}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\text{Substitute:}\\\\d=\sqrt{(2-(-2))^2+(0-3)^2}=\sqrt{4^2+(-3)^2}=\sqrt{16+9}\\\\=\sqrt{25}=5\\\\Answer:\ \boxed{(x-3)^2+(y-(-2))^2=5^2\to\boxed{(x-3)^2+(y+2)^2=25}}$

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What value of g makes the equation 2g+6-14g=-6(g-5) true?

Sergeu [11.5K]

Answer:

g=-4

Step-by-step explanation:

2g+6-14g=-6(g-5)

-12g+6=-6g+30

-6g+6=30

-6g=24

g=-4

8 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.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

$f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0$

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0

3 years ago

A loan of $900 will be repaid in 36 installment payments of $30.22 each. What the finance charge for the loan?

zimovet [89]

First you need to multiply 36 *$30.22=$1087.92
This gives you the total amount paid

Then subtract 1087.92-900=187.92
This gives you the amount of money you paid in finance charges.

3 0

3 years ago

Find a irrational number that is between 5.2 and 5.5.

VladimirAG [237]

Answer:

5.35

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

How do I graph this someone plz tell me.

hodyreva [135]

Easy. your starting point is going to be at -2 on the y-axis. then from that point (-2) you're going to go down 1 and then over 2. then draw a straight line through those two points. The line should be a negative line.

7 0

3 years ago