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Readme [11.4K]
2 years ago
9

PLEASE I NEED HELP!!! NO LINKS!!

Mathematics
1 answer:
9966 [12]2 years ago
4 0

Answer:

(5)2

Step-by-step explanation:

As we know that the equation of circle with center at (h,k) ad radius r is given as-

(x−h)

2

+(y−k)

2

=r

2

Given that the centre of circle is (1,4) and radius is 5.

Therefore the equation of circle is-

(x−1)

2

+(y−4)

2

=(5)

2

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Need a little help with one please
Pani-rosa [81]

Answer:

That would be the side-side-side (SSS) postulate, which states that if all the sides of a triangle are in a fixed ratio to all the corresponding sides of another triangle, then the two triangles are said to be congruent.

Step-by-step explanation:

Looking at triangles ABC and DEF, we notice that:

\frac{AB}{DE} = \frac{AC}{DF}=\frac{BC}{EF}

since

\frac{3}{18} = \frac{6}{36}=\frac{7}{42}=\frac{1}{6}

Let me know if you have further questions.

4 0
3 years ago
What's 600 out of 100 as a mixed number?
larisa [96]
600/100 as a mixed number would be 6
5 0
3 years ago
Read 2 more answers
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
HELP ASAP! BRAINLIST IF CORRECT
joja [24]

Answer:

the last one if I'm correct

Step-by-step explanation:

but im not sure I always second guess

4 0
3 years ago
Read 2 more answers
.…: need this answers.....
Leto [7]
Sorry, cannot just give you answers without your input.
I'd suggest that you look up online or in your textbook each of the items under "Type of Boundary."  The results you'd get will likely help you fill in the rest of the boxes.
3 0
3 years ago
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