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never [62]
3 years ago
9

A box with 3 metal balls weighs 1840 kg.

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
3 0

Answer:

weight = 736 kg

Step-by-step explanation:

Let ball be denoted by b and box by x

Given that:

3b + x = 1840kg ----------- eq1

According to given scenario:

each ball is thrice heavy than box So,

b = 3x -------------- eq2

Putting value of b in eq1

3 (3x) + x = 1840 kg

9x + x = 1840 kg

10x = 1840 kg

Dividing both sides by 10

x = 184 kg

Now finding weight of betal ball

Putting value of x in eq 2

b = 3 (184)

b = 552 kg

So the weight of one ball and box is:

weight = ball + box

weight = 552kg + 184 kg

weight = 736 kg

i hope it will help you!

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Answer:

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Step-by-step explanation:

The total job is 1

Mike has done 1/3    and Dena has done 2/7 of the job   We still need to do x of the job to complete it

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We need to get a common denominator 3*7 = 21

Multiply each side by 21

21(1/3 + 2/7+ x)  = 1*21

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3 years ago
find an equation of the line through the given points. give the final answer in slope-intercept form (-2,2) (4,-1)​
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to get the equation of any straight line, we simply need two points off of it, let's use the points in the picture below.

(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-1}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{(-2)}}}\implies \cfrac{-3}{4+2}\implies \cfrac{-3}{6}\implies -\cfrac{1}{2}

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{(-2)}) \\\\\\ y-2 = -\cfrac{1}{2}(x+2)\implies y-2=-\cfrac{1}{2}x-1\implies y=-\cfrac{1}{2}x+1

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poppy
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3 years ago
Calculate:<br>a) QR <br>b) PS <br>c) The area of quadrilateral PQRS
Zigmanuir [339]

<em>Greetings from Brasil...</em>

For QR: we need to use the Sine Law in Any Triangle....

QS/SEN 72 = QR/SEN 25

6/SEN 72 = QR/SEN 25

6,3 = QR/SEN 25

<h3>QR = 2,66</h3>

For PS: we need to use the Cosine Law in Any Triangle....

PS² = PQ² + QS² - 2.PQ.QS.COS Q

PS² = 7,4² + 6² - 2.(7,4).6.COS 34

PS² = 90,76 - 73,61

PS = √17,14

<h3>PS = 4,14</h3>

For area we use Heron's Formula 2x.....

for ΔQRS = A1:

A1 = √[P.(P - QR).(P - RS).(P - QS)]

<em>where P = (QR + RS + QS)/2</em>

<em>A1 = </em>√[P.(P - QR).(P - RS).(P - QS)]

<em>      </em><em>RS = 6,26 </em><em>(using RS/SEN 97 = QS/SEN 72)</em>

P = (2,66 + 6,26 + 6)/2

P = 14,92/2 ⇒ P = 7,46

A1 = √[7,46.(7,46 - 2,66).(7,46 - 6,26).(7,46 - 6)]

A1 = 7,92

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A2 = √[P.(P - PQ).(P - PS).(P - QS)]

P = (7,4 + 4,14 + 6)/2 = 8,77

A2 = √[8,77.(8,77 - 7,4).(8,77 - 4,14).(8,77 - 6)]

A2 = 12,41

Total Area = A1 + A2

Total Area = 7,92 + 12,41

<h3>Total Area = 20,33</h3>

<em>see more:</em>

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= mA'B' / mAB

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Your choice of scale-factor = 3 is correct! Well done!

6 0
3 years ago
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