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3 years ago

12 people can wash 40 cars in 12 days. how many days will it take 4 people to wash 30 cars

1 answer:

7 0

First off, let's keep in mind we're looking for "days worth", how many days does it take to do an amount of cars.

so 12 folks can do in 12 days 40 cars, since each person is working every one of those 12 days, then each of those persons is working 12 days, now, there are 12 folks, that means the combined amount of days will be 12+12+12+12+12+12+12+12+12+12+12+12, or 12*12, 144 days worth, it takes them to do 40 cars.

so, to the same folks, how many days will it take them to do 30 cars?

$\bf \stackrel{12~folks}{\begin{array}{ccll}
days&cars\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
144&40\\
x&30
\end{array}}\implies \cfrac{144}{x}=\cfrac{40}{30}\implies \cfrac{144\cdot 30}{40}=x\implies 108=x$

that's 108 days worth for doing 30 cars.

for 12 folks, that'd be 108/12 or 9 days, 12 folks can do it in 9 days.

what about 4 folks only? 108/4 or 27 days.

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In ΔQRS, the measure of ∠S=90°, the measure of ∠R=29°, and SQ = 5.8 feet. Find the length of RS to the nearest tenth of a foot.

taurus [48]

Answer:

10.5 ft

Step-by-step explanation:

In ΔQRS,

m∠S=90°
m∠R=29°,
SQ = 5.8 feet.

See the diagram below.

The triangle is a Right Triangle,

Using trigonometric ratios,

$Tan R=\dfrac{|SQ|}{|RS|} \\Tan 29^\circ=\dfrac{5.8}{|RS|}\\$Cross multiply$\\|RS|XTan 29^\circ=5.8\\|RS|=\dfrac{5.8}{Tan 29^\circ}\\$=10.5 ft (to the nearest tenth of a foot)$

7 0

3 years ago

Thank you very much anh you

Neko [114]

Hello from MrBillDoesMath!

Answer:

See discussion below

Discussion:

The sum of the interior angles of a polygon of n sides is (n-2)* 180. As the polygon in #4 has 7 sides , the sum of the interior angles is (7-2)*180 = 5 * 180 = 900.

We can also compute the sum of the interior angles by adding up each of the angles (expressed in terms of x) shown in #4. Starting at the top of the polygon and proceeding counterclockwise gives:

900 = (8x +34) + (10x+21)+(9x+30)+(7x+45)+(5x+44)+(12x+13)+(6x+29) =>

(8x + 10x+9x+7x+5x+12x+6x) + (34+21+30+45+44+13+29) =

57x + 216

This simplifies to

900 = 57x + 216 => subtract 216 from both sides

900-216 = 57x => 900-216= 684

684= 57x => divide both sides by 57

684/57 = x => as 684/57 = 12

x = 12

The diagram has makes repeated use of the same variable for vertices. That is, the diagram shows 4 P's, 1 Q, 2 R's, 0 S's, and 1 T making it impossible to determine the values of some angles. For example, consider m RST -- whatever that means.

#5 uses the same ideas but the vertices are properly labeled (I'll leave the grunt of determining the individual angles to you but here are the main points).

Sum of interior angles = (7 -2) * 180 = 5 * 180 = 900.

900 = (8x+62) + (7x+66)+(5x+83)+(6x+55)+(10x+42)+(8x+45)+(4x+67) =>

900 = (8x + 7x +5x+6x+10x+8x+4x) + (62+66+83+55+42+45+67)

900= 48x + 420 =>

900 -420 = 48x => subtract 420

480 = 48x => divide by 48

x = 480/48 = 10

Thank you,

MrB

6 0

3 years ago

A 4-lb. force acting in the direction of (vector) 4,-2 moves an object just over 7ft from point (0,4) to (5,-1). Find the work d

Tcecarenko [31]

To solve this problem, we have to find the net displacement and the net force and the multiply the dot product together and get the work done.

The work done on moving the object is 27ft*lbs

<h3>Work done in moving the object from point A to point B</h3>

To find the work done on this object, let's find the net force on the object.

Data;

force = 4lb
direction = 4, -2
displacement = 7ft
direction = (0, 4) to (5,1)

The unit vector of the force is

$\sqrt{4^2 +(-2)^2} =\sqrt{16 + 4} = \sqrt{20}$

$\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} }$

The net force acting on the object is

$F = 4(\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} })\\F= (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } )$

The displacement on the object is 7ft through (0,4) to (5, -1)

The unit vector on displacement is

$\sqrt{5^2 + (-1-4)^2} = \sqrt{25+25} = \sqrt{50}$

$\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }$

The net displacement will be

$7(\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }) = \frac{35}{\sqrt{50} }, \frac{-35}{50}$

The work done will be F.d

$w = f. d \\$

$w = (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } ) * \frac{35}{\sqrt{50} }, \frac{-35}{50}\\w = 17.71+ 8.854\\w = 26.567 = 27ft*lbs$

The work done on moving the object is 27ft*lbs

Learn more on work done on an object here;

brainly.com/question/26152883

#SPJ1

6 0

2 years ago

Determine the number of possible solutions for a triangle with B=37 degrees, a=32, b=27

vladimir1956 [14]

Answer:

Two possible solutions

Step-by-step explanation:

we know that

Applying the law of sines

$\frac{a}{sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}$

we have

$a=32\ units$

$b=27\ units$

$B=37\°$

step 1

Find the measure of angle A

$\frac{a}{sin(A)}=\frac{b}{Sin(B)}$

substitute the values

$\frac{32}{sin(A)}=\frac{27}{Sin(37\°)}$

$sin(A)=(32)Sin(37\°)/27=0.71326$

$A=arcsin(0.71326)=45.5\°$

The measure of angle A could have two measures

the first measure-------> $A=45.5\°$

the second measure -----> $A=180\°-45.5\°=134.5\°$

step 2

Find the first measure of angle C

Remember that the sum of the internal angles of a triangle must be equal to $180\°$

$A+B+C=180\°$

substitute the values

$A=45.5\°$

$B=37\°$

$45.5\°+37\°+C=180\°$

$C=180\°-(45.5\°+37\°)=97.5\°$

step 3

Find the first length of side c

$\frac{a}{sin(A)}=\frac{c}{Sin(C)}$

substitute the values

$\frac{32}{sin(37\°)}=\frac{c}{Sin(97.5\°)}$

$c=Sin(97.5\°)\frac{32}{sin(37\°)}=52.7\ units$

therefore

the measures for the first solution of the triangle are

$A=45.5\°$ , $a=32\ units$

$B=37\°$ , $b=27\ units$

$C=97.5\°$ , $b=52.7\ units$

step 4

Find the second measure of angle C with the second measure of angle A

Remember that the sum of the internal angles of a triangle must be equal to $180\°$

$A+B+C=180\°$

substitute the values

$A=134.5\°$

$B=37\°$

$134.5\°+37\°+C=180\°$

$C=180\°-(134.5\°+37\°)=8.5\°$

step 5

Find the second length of side c

$\frac{a}{sin(A)}=\frac{c}{Sin(C)}$

substitute the values

$\frac{32}{sin(37\°)}=\frac{c}{Sin(8.5\°)}$

$c=Sin(8.5\°)\frac{32}{sin(37\°)}=7.9\ units$

therefore

the measures for the second solution of the triangle are

$A=45.5\°$ , $a=32\ units$

$B=37\°$ , $b=27\ units$

$C=8.5\°$ , $b=7.9\ units$

6 0

3 years ago

Find the measure of arc IK if theta equals 58°

timurjin [86]

Check the picture below.

8 0

2 years ago