Can you help me with 10-13 please? I don’t really understand it

In order to find the area for a strange shape, you divide it up into simpler figures. In this case, you already know that it can be divided into 4 triangles, so find the area of each triangle and add them up. Hope this helped!

6 0

4 years ago

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A. Rewrite the function y = x2 – 14x + 58 in vertex form by completing the square. Show your work. B. Rewrite the function y = –

marysya [2.9K]

Answer:

A. y = (x -7)² +9

B. y = -(x +3)² -11

C. minimum, (7, 9)

D. maximum, (-3, -11)

Step-by-step explanation:

The general steps to rewriting the function in vertex form are ...

1. factor out the leading coefficient (if not 1) from the first two terms

2. add inside parentheses the square of half the x coefficient; subtract the same amount outside parentheses. The factor in front of the parentheses must be taken into account when adding the same amount outside.

3. rewrite the content of parentheses as a square, collect terms outside parentheses.

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A. y = x² -14x +58

y = (x² -14x) +58 . . . . . . . . . . step 1

y = (x² -14x +49) +58 -49 . . . step 2

y = (x -7)² +9 . . . . . . . . . . . . . step 3

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B. y = -x² -6x -20

y = -(x² +6x) -20

y = -(x² +6x +9) -20 -(-9)

y = -(x +3)² -11

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C. In vertex form, the function for vertical scale factor "a" and vertex (h, k) looks like ...

y = a(x -h)² +k

When a < 0, the vertex is a maximum (the graph opens downward); when a > 0, the vertex is a minimum (the graph opens upward).

The function of part A has a minimum at the vertex (7, 9). 9 is the minimum value.

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D. The function of part B has a maximum at the vertex (-3, -11). -11 is the maximum value.

8 0

3 years ago

=

const2013 [10]

The equation in the given standard form is $(x+10)^2+(y+6)^2=5^2$

<h3>Equation of a circle</h3>

The standard equation of a circle is expressed as:

$(x-a)^2+(y-b)^2=r^2$

Given the equation $x^2 + y^2 + 20x + 12y + 111 =0$

Group the function to have:

$(x^2 +20x)+(y^2+ 12y) + 111 =0$

Complete the squares to have:

$(x^2 +20x)+(y^2+ 12y) + 111 =0\\(x^2+20x+10^2-10^2)+(y^2+12y+6^2-6^2)+111=0\\(x+10)^2+(y+6)^2 -100-36+111=0\\(x+10)^2+(y+6)^2-25=0\\(x+10)^2+(y+6)^2=25\\(x+10)^2+(y+6)^2=5^2$

Hence the equation in the given standard form is $(x+10)^2+(y+6)^2=5^2$

Learn more on equation of a circle here: brainly.com/question/1506955

4 0

3 years ago