I must confess that I was about to pass this question by, but I was captured by the respectful and dignified way in which you asked for help.
A careful reading of the problem gives you two equations in two unknowns, which you can then solve as simultaneous equations. Here's how it looks:
Call 'C' the price of the senior <u>C</u>itizen ticket.
Call 'S' the price of the <u>S</u>tudent ticket.
On the first night . . . 10 C + 12 S = 208
On the second night . . . 8C + 3 S = 74
Those are your two simultaneous equations. Now the idea is to multiply or divide each side of one equation in such a way that when you add or subtract it from the other equation, one of the variables will become a zero quantity ... you'll be left with an equation in one variable, which you can easily solve. THEN, knowing the value of one variable, you can put it back into one of the original equations,and find the value of the other variable.
This all sounds more complicated than it is. Here's how it goes:
We have . . .
10 C + 12 S = 208 and
8C + 3 S = 74
I'm going to multiply each side of the second equation by 4, and then write it under the first one:
10 C + 12 S = 208
32 C + 12 S = 296
Now, subtract the lower equation from the upper one, and you get . . .
- 22 C + 0 = - 88
Divide each side of this one by -22 and you have <em>C = $4.00</em> .
THAT's what you need, to blow the whole problem wide open. Knowing
the value of 'C', let's substitute it into the equation for the first night:
10 C + 12 S = 208
10(4) + 12 S = 208
40 + 12 S = 208
Subtract 40 from each side : 12 S = 168
Divide each side by 12 : <em>S =</em><em> $ 14.00 </em>.
Finally, as we look over our results, and see that Students have to pay $14 to see the show but Seniors can get in for only $4 , we reflect on this ... or at least I do ... and realize that getting old is not necessarily all bad.
nwere ike nwa osisi poop pusi ibu onu nkita meow n'anya papa 46721
Answer:
(four being the slope)
Step-by-step explanation:
So lets use
ΔY/ΔX
or the y2-y1 over x2-x1
Therefore, the slope is 4
Answer:
$17
Step-by-step explanation:
You need to find 20 percent. so find 10 percent or divide by ten then multiply that by two
Since the height of an equilateral triangle in terms of its side s is s√3/2, the height of the triangle is 6√3/2 = 3√3 and so the area is (1/2)(6)(3√3) = 9√3.
<span>If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3√3 - h. </span>
<span>Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have: </span>
<span>(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle) </span>
<span>==> x/6 = (3√3 - h)/(3√3) </span>
<span>==> x = (6√3 - 2h)/√3 </span>
<span>Thus, the area of the upper triangle is: </span>
<span>A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3). </span>
<span>(Made a dumb mistake about the height here for some reason) </span>
<span>Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve: </span>
<span>[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2 </span>
<span>==> (6√3 - 2h)(3√3 - h) = 27 </span>
<span>==> 54 - 6h√3 - 6h√3 + 2h^2 = 27 </span>
<span>==> 2h^2 - 12h√3 + 27 = 0. </span>
<span>Solving with the Quadratic Formula gives: </span>
<span>h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units. </span>
<span>Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2. </span>
<span>Therefore, the line should be ==> (6√3 - 3√6)/2 units from the base. </span>
<span>I hope this helps! ^^ Brainliest Please?</span><span>
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