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inn [45]
3 years ago
15

Suppose that the maximum weight that a certain type of rectangular beam can support varies inversely as it’s length and jointly

as its width and the square of its height. Suppose also that a beam 5 inches wide, 2 inches high and 10 feet long can support a maximum weight of 8 tons. What is the maximum weight that could be supported by a beam that is 6 inches wide, 2 inches high, and 24 feet long
Mathematics
1 answer:
Gre4nikov [31]3 years ago
8 0

Way: W = k * w  * h^2/ L

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27 millimeters multiplied by 1 cm/10 mm
Basile [38]
27 millimeters multiplied by 1 cm/10 mm= 2.7 centimeters

6 0
3 years ago
Pls help I’ll brainlest solve it like this
sergey [27]

\frac{3}{5} t = 6

t = 6 \div  \frac{3}{5}

t = 6 \times  \frac{5}{3}

t = 2 \times 5

t = 10

____________________________

\frac{1}{4} a =  \frac{14}{15}

a =  \frac{14}{15}  \div  \frac{1}{4}

a =  \frac{14}{15}  \times 4

a =  \frac{56}{15}

7 0
3 years ago
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How would you write x+3y=3 in y=mx+b format
Triss [41]
X+3y=3
First add subtract the x
3y=-x+3
Then divide 3
Y=-1/3x+1
8 0
3 years ago
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Solve for n.<br> n + 1 = 4(n - 8)<br> n = 1<br> T<br> n = 8<br> n = 11<br> n = 16
Komok [63]

Answer:

n=11

Step-by-step explanation:

n+1=4(n-8)

11+1=4(11-8)

12=4(3)

12=12

3 0
3 years ago
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What is the solution set for this linear-quadratic system of equations?
Lostsunrise [7]
Y - x - 3 = 0
y = x + 3

y = x^2 - x - 12
x + 3 = x^2 - x - 12
x^2 - x - 12 - x - 3 = 0
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0

x - 5 = 0     y = x + 3
x = 5          y = 5 + 3
                  y = 8

x + 3 = 0     y = x + 3
x = -3          y = -3 + 3
                   y = 0

so ur solutions are : (5,8) and (-3,0)
3 0
3 years ago
Read 2 more answers
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