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Vedmedyk [2.9K]
3 years ago
9

A soccer ball with a diameter of 8.6 inches is shipped in a box that is a square prism and has a side length of 9.5 inches.How m

uch volume is available to be filled with packing material the shipping company wants the box completely full? Round your answer to the nearest tenth.
1,190.2 in^3
774.2 in^3
1,805.6 in^3
5,115.8 in^3
524.5 in^3

Mathematics
1 answer:
allochka39001 [22]3 years ago
7 0

Answer:

524.5 in³

Step-by-step explanation:

First, we can find the volume of the box:

V = s³

V = 9.5³

V = 857.38 in³

Then, we can find the volume of the soccer ball:

V = (4/3)\pir³

V = (4/3)\pi(4.3)³

V = 333.04 in³

Then, subtract the volume of the ball from the volume of the box:

857.4 - 333.0 = 524.4

So, this is closest to the answer 524.5 in³

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Will mark brainliest for the correct answer!
romanna [79]

Part (a)

Focus on triangle PSQ. We have

angle P = 52

side PQ = 6.8

side SQ = 5.4

Use of the law of sines to determine angle S

sin(S)/PQ = sin(P)/SQ

sin(S)/(6.8) = sin(52)/(5.4)

sin(S) = 6.8*sin(52)/(5.4)

sin(S) = 0.99230983787513

S = arcsin(0.99230983787513)

S = 82.889762826274

Which is approximate

------------

Use this to find angle Q. Again we're only focusing on triangle PSQ.

P+S+Q = 180

Q = 180-P-S

Q = 180-52-82.889762826274

Q = 45.110237173726

Which is also approximate.

A more specific name for this angle is angle PQS, which will be useful later in part (b).

------------

Now find the area of triangle PSQ

area of triangle = 0.5*(side1)*(side2)*sin(included angle)

area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)

area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)

area of triangle PSQ = 13.0074347717966

------------

Next we'll use the fact that RS:SP is 2:1.

This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.

We found

area of triangle PSQ = 13.0074347717966

So,

area of triangle RSQ = 2*(area of triangle PSQ)

area of triangle RSQ = 2*13.0074347717966

area of triangle RSQ = 26.0148695435932

------------

We're onto the last step. Add up the smaller triangular areas we found

area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)

area of triangle PQR = (13.0074347717966)+(26.0148695435932)

area of triangle PQR = 39.0223043153899

------------

<h3>Answer: 39.0223043153899</h3>

This value is approximate. Round however you need to.

===========================================

Part (b)

Focus on triangle PSQ. Let's find the length of PS.

We'll use the value of angle Q to determine this length.

We'll use the law of sines

sin(Q)/(PS) = sin(P)/(SQ)

sin(45.110237173726)/(PS) = sin(52)/(5.4)

5.4*sin(45.110237173726) = PS*sin(52)

PS = 5.4*sin(45.110237173726)/sin(52)

PS = 4.8549034284642

Because RS is twice as long as PS, we know that

RS = 2*PS = 2*4.8549034284642 = 9.7098068569284

So,

PR = RS+PS

PR = 9.7098068569284 + 4.8549034284642

PR = 14.5647102853927

-------------

Next we use the law of cosines to find RQ

Focus on triangle PQR

c^2 = a^2 + b^2 - 2ab*cos(C)

(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)

(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)

(RQ)^2 = 136.420523798282

RQ = sqrt(136.420523798282)

RQ = 11.6799196828694

--------------

We'll use the law of sines to find angle R of triangle PQR

sin(R)/PQ = sin(P)/RQ

sin(R)/6.8 = sin(52)/11.6799196828694

sin(R) = 6.8*sin(52)/11.6799196828694

sin(R) = 0.4587765387107

R = arcsin(0.4587765387107)

R = 27.3081879220073

--------------

This leads to

P+Q+R = 180

Q = 180-P-R

Q = 180-52-27.3081879220073

Q = 100.691812077992

This is the measure of angle PQR

subtract off angle PQS found back in part (a)

angle SQR = (anglePQR) - (anglePQS)

angle SQR = (100.691812077992) - (45.110237173726)

angle SQR = 55.581574904266

--------------

<h3>Answer: 55.581574904266</h3>

This value is approximate. Round however you need to.

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Line AC is tangent to circle O at point C<br> What is the measure of OAC?
Nuetrik [128]
AC is a tangent so by definition, it touches the circle at exactly one point (point C) and forms a right angle at the tangency point. So angle ACO is 90 degrees

The remaining angle OAC must be 45 degrees because we need to have all three angles add to 180
45+45+90 = 90+90 = 180

Alternatively you can solve algebraically like so
(angle OAC) + (angle OCA) + (angle COA) = 180
(angle OAC) + (90 degrees) + (45 degrees) = 180
(angle OAC) + 90+45 = 180
(angle OAC) + 135 = 180
(angle OAC) + 135 - 135 = 180 - 135
angle OAC = 45 degrees

Side Note: Triangle OCA is an isosceles right triangle. It is of the template 45-45-90. 
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Kalon has $175 and needs to save at least $700 for a new computer. If he can save $35 per week, what is the minimum number of we
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700-175 = 525

525 divided by 35 = 15

It will take 15 weeks  


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3 years ago
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