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algol13
2 years ago
12

Bob has some 10 lb weights and some 3 lb weights. Together, all his weights add up to 50 lbs. If x represents the number of 3 lb

weights and y represents the number of 10 lb weights, which equation can be used to find the number of each type of weight Bob has?
Mathematics
1 answer:
Romashka [77]2 years ago
8 0
Total weight = 50 lb
x = number of 3-lb weights
y = number of 10-lb weights

weight of 3-lb weights = 3x
weight of 10-lb weights = 10y
total weight = 3x + 10y

equation
3x + 10y = 50
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Write the inequalities that are given by the :

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The inequality 7x + 4y ≤ 56 is graphed by drawing the line 7x + 4y = 56 and shading the region below that line.

The line 7x + 4y = 56 has these x and y intercepts:

y-intercept: x =0 => 4y = 56 => y = 56/4 => 14 => point (0,14)

x-intercept => y = 0 => 7x = 56 => x = 56/7= 8  => point (8,0)

So, the line passes through the poins (0,8) and (14,0) and the solution region is below that line.

Also, you know that x and y are restricted to be positive or zero =>

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Line joining ordered pairs 0, 14 and 8, 0. Shade the portion of the graph below this line which lies within the first quadrant
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2 years ago
Find the first three iterates of the function f(z) = 2z + (3 - 2i) with an initial value of
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Answer:

a.  5 + 2i, 13 + 2i, 29 + 2i

Step-by-step explanation:

We'll use the formula  f(z) = 2z + (3 - 2i)  for each iteration. The output of the first iteration will be come the input of the second iteration, and so on.

So, we start with z0 = 1 + 2i and we plug that into the base equation:

z0 = 1 + 2i ==> f(z) = 2(1 + 2i) + 3 - 2i = 2 + 4i + 3 - 2i = 5 + 2i

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z2 = 13 + 2i ==> f(z) = 2(13 + 2i) + 3 - 2i = 26 + 4i  + 3 - 2i = 29 + 2i

z3 = 29 + 2i

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2 years ago
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Answer:

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Step-by-step explanation:

Let the origin be the point 1 from where Ann start walking.

Ann walks 80 meters on a straight line 33° north of the east starting at point 1 as shown in figure below,

Resolving into the vectors, the vertical component will be 80Sin33° and Horizontal component will be 80Cos33° as shown in figure (2)

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Thus, Anna's walk  as a vector representation is 80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}

 




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