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Natasha_Volkova [10]
3 years ago
14

Which best defines a circle? A. A line segment joining the center to a point on the circumference. B. A line segment joining two

points on a curve. C. The set of all points that are the same distance from a given point called the center. D. A line segment joining two points on the circumference that passes through the center.
Mathematics
2 answers:
VARVARA [1.3K]3 years ago
6 0
I think it's D it's the only one that make sense to me
inna [77]3 years ago
5 0
I am not completely sure but I think it might be D, hope this helps!!
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Please help me find the area of the composite figure :)
kykrilka [37]
Find the area of both shapes in the composite figure and add them together.

A of rectangle is 6 x 12
= 72

A of triangle is 1/2 x 8 x 5
=20

Add them together to get 92 inches^2

Hope this helps!!
6 0
3 years ago
In ΔRST, the measure of ∠T=90°, the measure of ∠R=56°, and ST = 73 feet. Find the length of TR to the nearest tenth of a foot.
kramer

Answer:

49.2 feet

Step-by-step explanation:

x=73/Tan56=49.2391=49.2

5 0
3 years ago
A regular section of land made up of wheat farms has a length of 5x10^4 and the width of 6x10^3 what is the area of the length i
Fudgin [204]

Answer:

The Area of land is 30 × 10^{7} meters² .

Step-by-step explanation:

Given as :

The length of rectangular wheat farm land = L = 5 × 10^{4}  meter

The width of rectangular wheat farm land = w = 6 × 10^{3}  meter

Let The Area of land = A meters²

Now, According to question

∵ Land is in rectangular shape

And we know, Area of rectangle = length × width

∴ Area of land =  Area of rectangle = length × width

Or, A = L × w

Or, A = 5 × 10^{4}  meter × 6 × 10^{3}  meter

Or , A = 30 × 10^{4+3}

i.e A = 30 × 10^{7}  meters²

So, The Area of land = A  = 30 × 10^{7} meters²

Hence,The Area of land is 30 × 10^{7} meters² . Answer

8 0
4 years 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

so

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
Helpppppppppppppppppppp please...................
vlabodo [156]

Answer:

C and D

I am pretty sure

5 0
3 years ago
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