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3 years ago

Find the length of WRIf M <WCR = 49 and CR = 7. What is WR

Mathematics

2 answers:

UkoKoshka [18]3 years ago

4 0

We know that
if ∠WCR=49°
then
measure arc WR=49°-------> by central angle

circumference C=2*pi*r
for r=7 in
C=2*pi*7-----> 43.96 in

if 360°(full circle) has a length of----------> 43.96 in
49°---------------> x
x=49*43.96/360-----> x=5.98 in

alternative method
applying the formula
L=(∅/360°)*2*pi*r
where
∅ is the angle in degrees
r is the radius
L=(49°/360°)*2*pi*7------> L=5.98 in

the answer is
5.98 in

dem82 [27]3 years ago

4 0

Answer:

5.99in

Step-by-step explanation:

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Read 2 more answers

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

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3 years ago