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3 years ago

For a button to fit through its buttonhole, the hole needs to be the size of the button's diameter. What

Mathematics

2 answers:

alexira [117]3 years ago

6 0

Answer is A.2.35cm
c=pid
circumstances is 7.38 cm so
7.38=3.14d
divide both sides by 3.14
7.38/3.14=3.14x/3.14
2.35=×

Zinaida [17]3 years ago

4 0

IT's A. 2.35 centimeters

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Find the ratio of 5 degree and (pi/6)^c

Nataly_w [17]

Lemme just say this real quick, Ratio

3 0

2 years ago

In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC a

nirvana33 [79]

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

$\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ}$ ;

$\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ}$ ;

$\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q$ .

Also,

$\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P$ in right triangle PBM.

Thus $\rm \angle{P\hat{B}M} = \angle C\hat{M}Q$ .

Additionally $\rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}$ .

Therefore $\rm \triangle PBM \sim \triangle MCQ$ .

$\rm \displaystyle BC = 2\;MC$ for M is the midpoint of segment BC.

$\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC$ .

$\rm \triangle PBM \sim \triangle MCQ$ implies that

$\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3$ .

8 0

2 years ago

-6+ (-19) =<br> Evaluate

artcher [175]

The answer if you need helo with math like this come to me -25

5 0

2 years ago

What is the measure of RPQ

patriot [66]

Answer:

239

Step-by-step explanation:

SP is the diameter so Arc RP is supplementary to Arc SR

$31 + x = 180$

$x = 149$

Arc PQ is supplementary to Arc SQ so

$90 + x = 180$

$x = 90$

Arc Addition:

$149 + 90 = 239$

8 0

2 years ago

Sally lives in City A and works for the department of transportation. She needs to inspect the weigh stations in each of four di

galben [10]

Answer:

She can visit the cities in 192 ways

Step-by-step explanation:

If Sally starts from city A and after going to 4 cities she returns to home,

then that can be in 4! = 24 ways. (permutation of 4 distinct objects)

If between the time she visits cities, she comes home once,

then cities can be visited in,

$4! \times 3_{C_{1}}$ = 72 ways. (permutation of 4 distinct objects and choosing 1 gap among the 3 )

If between the time she visits cities, she comes home twice,

then cities can be visited in,

$4! \times 3_{C_{2}}$ =72 ways. (permutation of 4 distinct objects and finding 2 gaps among the 3)

If between the time she visits cities, she comes home thrice,

then cities can be visited in,

$4! \times 3_{C_{3}}$ = 24 ways. (permutation of 4 distinct objects and choosing 3 gaps among the 3)

Now, she can't come home more than thrice in between the time she visits cities (since there are 4 cities to visit only)

So, the number of ways she can visit cities = (24+ 72 + 72 + 24)

= 192

6 0

3 years ago