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likoan [24]
3 years ago
15

For which triangle can you find the unknown value of the variable using the law of sines?

Mathematics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

B

Step-by-step explanation:

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How to solve 9x-4 when x= 2/3
Readme [11.4K]

Answer:

2

Step-by-step explanation:

9(2/3)-4

2/3 of 9 is 6

6-4=2

3 0
3 years ago
Read 2 more answers
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



3 0
3 years ago
Read 2 more answers
Write an equivalent expression by applying the distributive property or drawing a diagram. You MUST show all work
Ipatiy [6.2K]
Here is the answer to the equation. The photo below is the work shown.

x^2-3x-18

3 0
3 years ago
Read 2 more answers
How many packs of pencils can you get if there is 1,440 pencils and 6 in each pack
mamaluj [8]

Divide 1440 by 6 to get 240 packs.

4 0
3 years ago
In order to solve the system of linear equations, the coefficients of one of the variables must be the same number but opposite
Mashutka [201]

Answer:

Step-by-step explanation:

because both equations do not have matching numbers for the x or y variable, and both equations are positive you are going to have to multiply each equation by a number  so that there will be at least one variable with the same number but with opposite signs.

it does not matter which variable you choose.

lets use y because 2 and 3 are smaller then 2 and 5.

so lets multiply the first equation by 2 in order to get y equal to 6.

2(2x)+2(3y)=(2)6

(do not forget to multiply what the equation is equal to also)

4x+6y=12

now for the second equation we need y to equal negative 6

-3(5x)+-3(2y)=-3(4)

-15x-6y=-12

now lets put the 2 new equations next to each other and see what we can cancel out

4x+6y=12

-15x-6y=-12

-11x=0

x=0

now plug 0 in for x and solve for y (it does not matter which of the 4 equations you choose to solve.

2(0)+3y=6

3y=6

y=2

so your answer is x=0, y=2

8 0
3 years ago
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