I need help with this question please
1 answer:
Answer:
w =
Step-by-step explanation:
We have t = 4w, the question is asking to rearrange the equation to make "w" the subject, meaning that we have to isolate the variable w.
When we have a number (usually known as a coefficient) next to a variable and want to get rid of it, we need to divide it by both sides. Therefore :
t = 4w
[]
Since "t" is an undefined variable and has no initial value, you cannot divide it by anything, therefore it'll stay as [t/4 = w]
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F(x) = 4x - 1
g(x) = 2x² + 3
1. (f + g)(x) = (4x - 1) + (2x² + 3)
(f + g)(x) = 2x² + 4x + (-1 + 3)
(f + g)(x) = 2x² + 4x + 2
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
2. (f - g)(x) = (4x + 1) - (2x² + 3)
(f - g)(x) = 4x + 1 - 2x² - 3
(f - g)(x) = -2x² + 4x + 1 - 3
(f - g)(x) = -2x² + 4x - 2
Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
(g - f)(x) = 2x² + 3 - 4x + 1
(g - f)(x) = 2x² - 4x + 3 + 1
(g - f)(x) = 2x² - 4x + 4
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
4. (f · g)(x) = (4x + 1)(2x² + 3)
(f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
(f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
(f · g)(x) = 8x³ + 12x + 2x² + 3
(f · g)(x) = 8x³ + 2x² + 12x + 3
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
5.
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)
6.
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4 4
x ≠ 0
(-∞, 0) ∨ (0, ∞)
g(x) = 2x² + 3
1. (f + g)(x) = (4x - 1) + (2x² + 3)
(f + g)(x) = 2x² + 4x + (-1 + 3)
(f + g)(x) = 2x² + 4x + 2
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
2. (f - g)(x) = (4x + 1) - (2x² + 3)
(f - g)(x) = 4x + 1 - 2x² - 3
(f - g)(x) = -2x² + 4x + 1 - 3
(f - g)(x) = -2x² + 4x - 2
Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
(g - f)(x) = 2x² + 3 - 4x + 1
(g - f)(x) = 2x² - 4x + 3 + 1
(g - f)(x) = 2x² - 4x + 4
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
4. (f · g)(x) = (4x + 1)(2x² + 3)
(f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
(f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
(f · g)(x) = 8x³ + 12x + 2x² + 3
(f · g)(x) = 8x³ + 2x² + 12x + 3
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
5.
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)
6.
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4 4
x ≠ 0
(-∞, 0) ∨ (0, ∞)
The complete question in the attached figure
we know that
(see the attached figure n 2 to understand the problem)
[the surface area of one prism]=2*[x*x]+2*[x*y]+2*[x*y]----> 2x²+4xy
[the surface area of the sculpture]=2*[5*x*y]+2*[3*x*x]+2*[3*x*y]--> 6x²+16xy
now
<span>JD says the surface area of the sculpture is 4 times the surface area of one prism
</span>[the surface area of the sculpture]=4*(2x²+4xy)---> 8x²+16xy
we compare the value that JD says with the real value
(8x²+16xy) > (6x²+16xy)
the value that JD says is <span>greater in comparison with the real value
</span>This is because <span>JD should also subtract the areas of eight hidden surfaces.
the answer is
</span>JD should also subtract the areas of eight hidden surfaces<span>
</span>
we know that
(see the attached figure n 2 to understand the problem)
[the surface area of one prism]=2*[x*x]+2*[x*y]+2*[x*y]----> 2x²+4xy
[the surface area of the sculpture]=2*[5*x*y]+2*[3*x*x]+2*[3*x*y]--> 6x²+16xy
now
<span>JD says the surface area of the sculpture is 4 times the surface area of one prism
</span>[the surface area of the sculpture]=4*(2x²+4xy)---> 8x²+16xy
we compare the value that JD says with the real value
(8x²+16xy) > (6x²+16xy)
the value that JD says is <span>greater in comparison with the real value
</span>This is because <span>JD should also subtract the areas of eight hidden surfaces.
the answer is
</span>JD should also subtract the areas of eight hidden surfaces<span>
</span>