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3 years ago

AB is tangent to the circle at B. M∠A = 27 and mBC=114 (The figure is not drawn to scale.)

Mathematics

1 answer:

JulsSmile [24]3 years ago

8 0

9514 1404 393

Answer:

a. x = 60

b. y = 93

Step-by-step explanation:

The relevant relations are ...

external angle A is half the difference of intercepted arcs BC and BD
inscribed angle y° is half the measure of intercepted arc CD
the sum of arcs of a circle is 360°

Using these relations, we have ...

A = (BC -x°)/2

x° = BC -2A = 114° -2(27°)

x° = 60°

y° = CD/2 = (360° -BC -BD)/2 = (360° -114° -60°)/2

y° = 93°

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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...

7 0

3 years ago

Please help I give you 30 points

valentinak56 [21]

Answer:

The last one, 14 3/4

Step-by-step explanation:

I hope this helped :)

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2 years ago

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Ymorist [56]

Answer:

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Step-by-step explanation:

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3 years ago

Can someone please help me? :(

Anit [1.1K]

Answer:

____________________________________

$\large{ \tt{ \star \: S \: O \: L \: U \: T \: I \: O \: N : }}$

$\large{ \boxed{ \tt{♡ \: VOLUME \: = l \: \times b \: \times h}}}$

$\longrightarrow \: 4 \frac{1}{3} \times 3 \frac{1}{3} \times 4$

$\longrightarrow \: \frac{13}{3} \times \frac{10}{3} \times 4$

$\longrightarrow \: \frac{540}{9}$

$\longrightarrow \tt \: 57 \frac{7}{9} \: ft ^{3}$

Our final answer is 57 7 / 9 ft^3 .

- HappY LearninG ^_________^

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motikmotik

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