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3 years ago

8

Find the perimeter of a rectangle measuring 513 mm by 712mm

2 answers:

SSSSS [86.1K]3 years ago

5 0

I think you have to add 513mm and 712mm to get the perimeter of your rectangle hope this helped

alexdok [17]3 years ago

5 0

<em>h x w = perimeter
h=513mm w=712mm

</em>513mm x 712mm = 365,256mm

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Nesterboy [21]

Answer: $230^{\circ$

Step-by-step explanation:

For convenience, I labeled some points as shown in the attached picture.

Also, I assume $\overline{AB}$ and $\overline{BC}$ are tangents to the circle.

$\overline{BO} \cong \overline{BO}$ (reflexive property)
$\overline{AB} \cong \overline{BC}$ (tangents drawn from a common external point are congruent)
$\angle BAO \cong \angle BCO$ (right angles are congruent)

Therefore, we know $\triangle ABO \cong \triangle CBO$ by HL.

Thus, by CPCTC,

$\angle AOB \cong \angle BOC \implies m\angle BOC=65^{\circ}\\\\\implies m\angle AOC=130^{\circ}$

This means the measure of minor arc AC is $130^{\circ}$ , and thus $b=360^{\circ}-130^{\circ}=\boxed{230^{\circ}}$

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2 years ago

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Sunny_sXe [5.5K]

Answer:

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Part 2) The vertex is the point $(1,8)$

Part 3) The equation of the axis of symmetry is $x=1$

Part 4) The vertex is a minimum

Part 5) The quadratic equation in standard form is $y=2x^{2}-4x+10$

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

if a > 0 then the parabola open upward (vertex is a minimum)

if a < 0 then the parabola open downward (vertex is a maximum)

The equation of the axis of symmetry of a vertical parabola is equal to the x-coordinate of the vertex

so

$x=h$

In this problem we have

$y=2(x-1)^{2}+8$ -----> this is the equation in vertex form of a vertical parabola

The value of $a=2$

so

a>0 then the parabola open upward (vertex is a minimum)

The vertex is the point $(1,8)$

so

$(h,k)=(1,8)$

The equation of the axis of symmetry is $x=1$

The equation of a vertical parabola in standard form is equal to

$y=ax^{2}+bx+c$

Convert vertex form in standard form

$y=2(x-1)^{2}+8$

$y=2(x^{2}-2x+1)+8$

$y=2x^{2}-4x+2+8$

$y=2x^{2}-4x+10$

see the attached figure to better understand the problem

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