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julia-pushkina [17]
3 years ago
8

Last question for the day.. can anyone help wit #14 and either #10-13

Mathematics
2 answers:
fiasKO [112]3 years ago
8 0
Question 14 I think it is 2/5
masha68 [24]3 years ago
3 0
10. (5,8)(-3,7)
      slope = (7 - 8) / (-3 - 5) = -1 / -8 = 1/8

11. (5,-2)(3,-2)...since both y values are the same, this line is a horizontal line with a 0 slope.

12. (-4,7)(8,-1)
      slope = (-1 - 7) / (8 - (-4) = -8 / (8 + 4) = -8/12 = - 2/3

13. (6,-3)(6,4)....since both x values are the same, this line is a vertical line with a slope that is undefined

14. -2/5
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For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



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Read 2 more answers
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Answer:

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Step-by-step explanation: is the answer

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WARRIOR [948]
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I need some help, with my algebra ​
Viktor [21]

Answer:

Hey there!

Perimeter: 24

Area: 24

36+64=x^2

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Perimeter: 10+6+8=24

Area: 1/2bh, 1/2(48)=24

Both the area and perimeter are 24.

Hope this helps :)

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Why don't you first try to use the cosine law to solve for an angle and then make use of the sin law to solve for the remaining angles.

Cosine law

C^2 = A^2 + B^2 - 2AB(cos C)
Solve for cos C, and then take the inverse of the trig ratio to solve for the angle.

Then set up a proportion like you have done using the sin law and solve for another angle. Knowing the sum of all angles in a triangle add up to 180 degrees, we can easily solve for the remaining angle.
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