X=2h, y=3k
Substitute these values into equations.
y+2x = 4 ------> 3k+2*2h=4 -----> 3k +4h =4
2/y - 3/2x = 1-----> 2/3k -3/(2*2h) = 1 ------> 2/3k - 3/4h =1
We have a system of equations now.
3k +4h =4 ------> 3k = 4-4h ( Substitute 3k in the 2nd equation.)
2/3k - 3/4h =1
2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0
4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0
(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
Numerator should be = 0
4h- 3(2-2h) - 4h(2-2h)=0
Denominator cannot be = 0
4h(2-2h)≠0
Solve equation for numerator=0
4h- 3(2-2h) - 4h(2-2h)=0
4h - 6+6h-8h+8h² =0
8h² +2h -6=0
4h² + h-3 =0
(4h-3)(h+1)=0
4h-3=0, h+1=0
h=3/4 or h=-1
Check which
4h(2-2h)≠0
1) h= 3/4 , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.
h=3/4, then 3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then 3k = 4-4*(-1) =8 , 3k=8, k=8/3
So,
if h=3/4, then k=1/3,
and if h=-1, then k=8/3 .
The slope of the line connecting two points (<em>a</em>, <em>b</em>) and (<em>c</em>, <em>d</em>) is
(<em>d</em> - <em>b</em>) / (<em>c</em> - <em>a</em>)
i.e. the change in the <em>y</em>-coordinate divided by the change in the <em>x</em>-coordinate. For a function <em>y</em> = <em>f(x)</em>, this slope is the slope of the secant line connecting the two points (<em>a</em>, <em>f(a)</em> ) and (<em>c</em>, <em>f(c)</em> ), and has a value of
(<em>f(c)</em> - <em>f(a)</em> ) / (<em>c</em> - <em>a</em>)
Here, we have
<em>f(x)</em> = <em>x</em> ²
so that
<em>f</em> (1) = 1² = 1
<em>f</em> (1.01) = 1.01² = 1.0201
Then the slope of the secant line is
(1.0201 - 1) / (1.01 - 1) = 0.0201 / 0.01 = 2.01
It is 4900mm^3 because there are 1000mm^3 in one 1cm^3.
If you do 10^3 (10 mm) it gives you 1000mm^3.
Hope this helps.
Answer:
Associative
Step-by-step explanation:
When three or more numbers are multiplied, the product is the same regardless of the way in which the numbers are grouped.
I think is C that what i think:;;:::::);:;:)
