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Evgesh-ka [11]
2 years ago
11

What is the radius of a circle in which a 30 degrees arc is 2 inches long

Mathematics
1 answer:
igomit [66]2 years ago
6 0

arc length = pi /180 * angle in degree * radius

2 = pi/180 * 30 *r

2 = pi/6 *r

multiply by 6/pi

12/pi =r

3.819718634 =r

Answer: r =3.82 inches


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Sal owns a clothing store that sells shorts and graphic T-shirts. He sells the
oee [108]

Answer:

You have a thing on there saying the answer is c. If you got the answer there is no point of asking it.

7 0
2 years ago
Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7
photoshop1234 [79]
Hope I can be of assistance!

3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \  \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}
3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \  3b^{-5}b^2c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  b^{-5}b^2=\:b^{2-5}=\:b^{-3}
3b^{-3}c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  c^{-4}c^7=\:c^{-4+7}=\:c^3
3b^{-3}c^3

\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \  b^{-3}=\frac{1}{b^3} \ \textgreater \  3\cdot \frac{1}{b^3}c^3

\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{1\cdot \:3c^3}{b^3}

Finally
\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \  \frac{3c^3}{b^3}

Hope this helps!
8 0
3 years ago
A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time).
likoan [24]

Answer:

(a). 72.9%.

(b). 13.6 hr.

Step-by-step explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this question/problem;

=> "A welder produces 7 welded assemblies during the first day on a new job, and the seventh assembly takes 45 minutes (unit time). "

=> The worker produces 10 welded assemblies on the second day, and the 10th assembly on the second day takes 30 minutes"

So, we will be making use of the Crawford learning curve model.

T(7) + 10 = T (17) = 30 min.

T(7) = T1(7)^b = 45.

T(17 ) = T1(17)^b = 30.

(T1) = 45/7^b = 30/17^b.

45/30 = 7^b/17^b = (7/17)^b.

1.5 = (0.41177)^b.

ln 1.5 = b ln 0.41177.

0.40547 = -0.8873 b.

b = - 0.45696.

=> 2^ -0.45696 = 0.7285.

= 72.9%.

(b). T1= 45/7^ - 045696 = 109.5 hr.

V(TT)(17) = 109.5 {(17.51^ - 0.45696 – 0.51^ - 0.45696) / (1 - 0.45696)} .

V(TT) (17) = 109.5 {(4.7317 - 0.6863) / 0.54304} .

= 815.7 min .

= 13.595 hr.

8 0
3 years ago
The slope of a line is 1, and the y-intercept is -1. What is the equation of the line written in slope-intercept form?
Scilla [17]
The slope-intercept form is:
y = mx + b
where m = slope, and b = y-intercept.
You need a slope of 1, so m = 1.
You need a y-intercept of -1, so b = -1.
Replace m with 1 and b with -1 in the slope intercept form to get

y = 1x + (-1)

which simplifies to

y = x - 1
3 0
3 years ago
What is the area, in yards, of<br>a box thats 38 feet wide by<br>15 feet<br> long?​
pshichka [43]

Answer:

570

step-by-step explanation:

assuming you're talking about a 2-dimensional rectangle,

Area = base x height

base = 38

width = 15

Area = 38 * 15

Area = 570 ft

570ft * (1 yard/3ft) = 190 yards

6 0
2 years ago
Read 2 more answers
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