Answer:
life span ................
Answer:
Mia had a profit
Step-by-step explanation:
Since every figure repr. an equil. triangle, each of the angles shown is 60 deg.
Thus, the height of one of the shaded triangles is s sin 60 deg, or s/sqrt(3).
The base of one such triangle is s/3.
Thus, the area of ONE shaded triangle is A = bh/2, which here is
(s/3)(s/sqrt(3) ) (s/3)(s)
A = ----------------------- = --------------------
2 2sqrt(3)
Rationalize the denom.: Mult numerator and denominator both by sqrt(3), and then mult the resulting numerator and den by 3:
sqrt(3) s^2/3 s^2*sqrt(3) s^2*sqrt(3)
----------- * ----------- = ----------------- = ----------------
sqrt(3) 2sqrt(3) 3*2*3 18
Unfortunately, this does not agree with any of the four answer choices.
Let's take a slightly different approach here:
Each (horizontal) base of each shaded triangle has length s/3. Each has height (s/3)sin(60 deg) = s*sqrt(3)/(3*2) = s*sqrt(3)/6.
[s/3]*[s*sqrt(3)]/6
Then each shaded triangle has area A = bh/2 = ----------------------------
2
and the total shaded area is three times that:
s^2 * sqrt(3) s^2*sqrt(3)
Total area = -------------------- = ------------------ (answer)
3*2 * 6 36
Answer:
3x²(4x-3)
Step by step explanation:
Factorise 12x³-9x²
first, find the HCF of the terms.
the HCF is 3x²
Then you can use the HCF to multiply all terms, each divided by the HCF.
3x²(12x³/3x²-9x²/3x²)
Dividing, we'll have
3x²(4x-3)
In "slope-intercept form"
y = mx +b
the value "m" is called the slope, and the value "b" is called the intercept.
There is another form for the equation of a line, called "point-slope form".
y = m(x -h) +k
where m is still the slope and (h, k) correspond to the (x, y) of the point.
If you write the equation of your line in this "point-slope form", it is easily manipulated to be in the "slope-intercept form".
Fill in
m = (-3/5)
h = -4
k = 0
y = (-3/5)(x -(-4)) +0
Now, you simplify this by using the distributive property.
y = (-3/5)x -(3/5)*4
y = (-3/5)x -12/5 . . . . . . . . . the desired equation
_____
Your understanding of math improves immensely when you become familiar with the terminology. A lot of the rest of it is pattern matching--identifying the parts of one expression that correspond to the parts of another one.
(You will see another version of the "point-slope form", but I find this one the easiest to use for manipulating the equation to other forms.)