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borishaifa [10]
3 years ago
11

What is the answer to 23,450 to the nearest thousands

Mathematics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

23,000

Step-by-step explanation:

brainliest

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In a circle of radius 10 cm, a sector has an area of 40 (pi) sq. cm. What is the degree measure of the arc of the sector?
Rashid [163]

Answer:

144°

Step-by-step explanation:

First, find the area of the circle, with the formula A = \pir²

Plug in 10 as the radius, and solve

A = \pir²

A = \pi(10²)

A = 100\pi

Using this, create a proportion that relates the area of the sector to the degree measure of the arc.

Let x represent the degree measure of the arc of the sector:

\frac{40\pi }{100\pi } = \frac{x}{360}

Cross multiply and solve for x:

100\pix = 14400\pi

x = 144

So, the degree measure of the sector arc is 144°

6 0
3 years ago
Read 2 more answers
Given that lines q and m are parallel and are cut by two
sashaice [31]

angle 1 = 180-110 = 70 degrees (linnear pair)

angle 2= 180-115 = 65 degrees (linnear pair)

angle 3 = angle 2 = 65 degrees (corresponding angles)

angle 4= 180 - angle 3 = 180-65 = 115 degrees (linnear pair)

(your teacher might not like this, depends on the cituation, but use 180 degrees in triangle to get 7)

angle 7 = 180 -angle1 -angle 2= 180-70-65= 45 degrees

angle 6 = angle 7 = 45 degrees ( vertical angles)

angle 8= 180- angle 7= 180-45 = 135 ( linnear pair)

angle 9 =angle 6 = 45 degrees (corresponding angles)

angle 5=angle 9=45 degrees (vertical angles)

7 0
3 years ago
PLEASE ANSWER THIS I BEG FOR YOUR MERCY!!!!!!!!!!!! if you dont know....dont answer
Andreyy89

Answer:

the answer is the second choice

Step-by-step explanation:

hopefully this helps :)

8 0
3 years ago
FIRST TO ANSWER GETS BRAINLIEST!
scZoUnD [109]

Answer:

\huge \boxed{ \red{ \boxed{c)78 \: units ^{2} }}}

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • circle
  • PEMDAS
<h3>tips and formulas:</h3>
  • \rm \:A_{shaded}=  \dfrac{\theta}{360}  \times \pi {r}^{2}
<h3>let's solve:</h3>
  1. \sf sustitute \: the \: value \: of \:  \theta \: and \: r :  \\ A_{shaded} =  \frac{140}{360}  \times \pi \times  {8}^{2}
  2. \sf simpify \: squre :  \\ A_{shaded} =  \frac{140}{360}  \times \pi \times  64
  3. \sf simpify \: fraction:  \\ A_{shaded} =  \frac{7}{18}  \times \pi \times  64
  4. \sf  reduce \: 64:  \\ A_{shaded} =  \frac{7}{ \cancel{ \: 18} \:  ^{9} }  \times \pi \times  \cancel{ 64}  \: ^{32}  \\ A_{shaded} =  \frac{7}{9} \times \pi \times 32\\ A_{shaded} =  \frac{224\pi}{9}

\large A_{shaded}\approx 78 \: square  \: units

8 0
3 years ago
Read 2 more answers
If p(q) = q^2 4q- 12 then what is p(-1)
guapka [62]

Answer:

\bold{p(q) = q^2 4q- 12 }  \\  \bold{p( - 1) =  {( - 1)}^{2} \times 4 \times ( - 1) - 12} \\  = \bold {1 \times ( - 4) - 12} \\  \bold{  = ( - 4) - 12} \\  \bold{ p( - 1)=  - 16}

5 0
2 years ago
Read 2 more answers
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