HELP PLEASE!! What is the constant of proportionality in the relationship?
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Answer:
(0.5,3.5)
Step-by-step explanation:
First, we can draw the image, as shown. The diagonals in the rectangle are the following lines:
from (-2,9) to (3,-2)
from (-5, 6) to (6,1)
To find where they intersect, we can start by making an equation for the lines. For an equation y=mx+b, m represents the slope and b represents the y intercept, or when x=0
For the first line, from (-2,9) to (3,-2), we can calculate the slope by calculating the change in y/change in x = (y₂-y₁)/(x₂-x₁). If (3,-2) is (x₂,y₂) and (-2,9) is (x₁,y₁), our slope is
(-2-9)/(3-(-2)) = -11/5
Therefore, our equation is
y= (-11/5)x + b
To solve for b, we can plug a point in, like (3,-2). Therefore,
-2=(-11/5)*3+b
-2=-33/5+b
-10/5=-33/5+b
add 33/5 to both sides to isolate b
23/5=b
Our equation for one diagonal is therefore y=(-11/5)x+23/5
For the second line, from (-5, 6) to (6,1), if (6,1) is (x₁,y₁) and (-5,6) is (x₂,y₂), the slope is (1-6)/(6-(-5)) = -5/11 . Plugging (6,1) into the equation y=(-5/11)x+b, we have
1=(-5/11)*6+b
11/11 = -30/11 + b
add 30/11 to both sides to isolate b
41/11 = b
our equation is
y = (-5/11) x + 41/11
Our two equations are thus
y = (-5/11) x + 41/11
y=(-11/5)x+23/5
To find where they intersect, we can set them equal to each other
(-11/5)x+23/5 = y = (-5/11) x + 41/11
(-11/5)x + 23/5 = (-5/11)x + 41/11
subtract 23/5 from both sides as well as add 5/11 to both sides to make one side have only x values and their coefficients
(-11/5)x + (5/11)x = 41/11-23/5
11*5 = 55, so 55 is one value we can use to make the denominators equal.
(-11*11/5*11)x+(5*5/11*5)x=(41*5/11*5)-(23*11/5*11)
(-121/55)x+(25/55)x = (205/55) - (253/55)
(-96/55)x = (-48/55)
multiply both sides by 55 to remove the denominators
-96x=-48
divide both sides by -96 to isolate x
x=-48/-96=0.5
plug x=0.5 into a diagonal to see the y value of the intersection
(-11/5)x + 23/5 = y = (-11/5)* 0.5 + 23/5 = 3.5
Given that <span>∆abc is isosceles and ab = bc, then m<bac = m<acb and m<hbc = 180 - m<bac - m<acb
Given that </span><span>m∠hbc = m∠bac +m∠bch, then m<hbc + m<bch = m<bac + 2m<bch
But m<hbc + m<bch = 90°, thus 90° = m<bac + 2m<bch
</span><span>Also, m∠bac + m∠ach = 90° ⇒ m<bac + 2m<bch = m<bac + m<ach ⇒ 2m<bch = m<ach
Since, Δabc is isosceles with ab = bc ⇒ m<bac = m<acb.
Also, m<acb = m<ach + m<bch ⇒ m<acb = 3m<bch = m<bac
</span><span><span>Since m<bch : m<ach = 1 : 2 ⇒ bh : ah = 1 : 2
Thus,
Given that ch = 84 cm, then
Now,
</span> </span>
Given that </span><span>m∠hbc = m∠bac +m∠bch, then m<hbc + m<bch = m<bac + 2m<bch
But m<hbc + m<bch = 90°, thus 90° = m<bac + 2m<bch
</span><span>Also, m∠bac + m∠ach = 90° ⇒ m<bac + 2m<bch = m<bac + m<ach ⇒ 2m<bch = m<ach
Since, Δabc is isosceles with ab = bc ⇒ m<bac = m<acb.
Also, m<acb = m<ach + m<bch ⇒ m<acb = 3m<bch = m<bac
</span><span><span>Since m<bch : m<ach = 1 : 2 ⇒ bh : ah = 1 : 2
Thus,
Given that ch = 84 cm, then
Now,
</span> </span>
The area of a shape is the amount of space it can occupy.
<em>The surface area of the larger prism is 4 times the surface area of the smaller prism. </em>
<em />
<em />
<em>Let the dimension of one side of the smaller prism be: l and w</em>
<em>Let the dimension of one side of the bigger prism be: L and W</em>
<em />
So, we have:
The area of these corresponding sides are:
---- the first
While, the area of the second is:
The ratio of these two sides are:
<em>This gives</em>
Cancel out common terms
Express as fractions
Also, we have:
This gives
Cross multiply
Hence, option (c) is true
Read more about areas at: