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3 years ago

In circle S with the measure of angle RST = 102 and RS = 6 units, find the length of arc RT. Round to the nearest hundredth

Mathematics

1 answer:

GREYUIT [131]3 years ago

3 0

Answer:

10.68

Step-by-step explanation:

\text{Arc Length}=

Arc Length=

\,\,\frac{\text{Central Angle}}{360}(\text{Circumference})

360

Central Angle

(Circumference)

RT=

\,\,\frac{102}{360}(2 \pi6)

360

102

(2π6)

RT=

\,\,10.68

10.68

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AB is a straight line workout the size of angle X

Hoochie [10]

The x value will be 109°. The straight line formed a 180° angle. Solving the equation yields the angle.

<h3>What are supplementary angles?</h3>

Supplementary angles are two angels whose sum is 180°. When a straight line intersects a line, two angles form on each of the sides of the considered straight line.

Those two-two angles are supplementary angles in two pairs. That is, if two supplementary angles are adjacent to each other, their exterior sides form a straight line.

The straight line formed a 180° angle. The resulting equation is as follows:

⇒x+42°+29°=180°

⇒x=109°

Hence, the value of the x will be 109°

The complete question is:

AB is a straight line.

Work out the size of angle x.

Not drawn accurately

42°

29°

To learn more about supplementary angles, refer to:

brainly.com/question/12919120

#SPJ1

4 0

1 year ago

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

Part c) How wide are the photo and frame together?

$(4+2x)=4+2(1.5)=7\ in$

5 0

3 years ago

What 7 coins can you use to get $0.77

Korvikt [17]

<span>3 quarters and 2 pennies.3 quarters= $0.75 2 pennies= $0.02= $0.77</span>

7 0

3 years ago

Which point is the reflection of point E across the y-axis ? What’s are it’s coordinates?

andriy [413]

Answer:

I need more info

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

-8x^2+4x+5=0 using quadratic formula

daser333 [38]

So you want to solve for x?

It would be nice if this would easily factor:
(-4x + 5)(2x +1) = 0    This will not work!

So you need to use the quadratic formula:
a = -8, b = 4, c = 5

$x = \frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$

x = (-4 +/- $\sqrt{16-4(-8)(5)}$ )/2(-8)
   = (-4 +/- $\sqrt{16 + 160}$ )/-16
   = (-4 + $\sqrt{176}$ )/-16
   = 1/4 - $\sqrt{11}$ /4

6 0

3 years ago