Found the problem:
Item Cost Sales Price
T-shirt 6 10
hats 4 7
Let x = t-shirts, y = hats
6x + 4y = 2,000
10x + 7y = 3,375
4y = 2000 - 6x
y = 2000/4 - 6x/4
y = 500 - 3x/2
10x + 7y = 3,375
10x + 7(500 - 3x/2) = 3,375
10x + 3500 - 21x/2 = 3,375
10x - 10.5x = 3,375 - 3500
-0.5x = -125
-0.5x / -0.5 = -125/-0.5
x = 250 shirts
They will need 250 shirts.
Answer:
option B and option D
0 and 2
Step-by-step explanation:
Given in the question two functions
f(x) = x² - 4x + 3
g(x) = -x² + 3
f(x) = g(x)
x² - 4x + 3 = - x² + 3
Rearrange the like terms, x terms to the left and constant term to the right.
x²+ x² - 4x = 3 - 3
2x² - 4x = 0
Divide by two
2x²/2 - 4x/2 = 0/2
x² - 2x = 0
Take x as a common term
x(x-2) = 0
so
x = 0
or
(x-2) = 0
x = 2
5).
and
6).
The volume of a sphere is
(4/3) (pi) (radius)³ .
In #5, the 'pi' is already there next to the answer window.
You just have to come up with the (4/3)(radius³).
Remember that the radius = 1/2 of the diameter.
7). The volume of a cylinder is
(pi) (radius²) (height) .
Divide the juice in the container by the volume of one can,
to get the number of cans he can fill.
8). The volume of a cone is
(1/3) (pi) (radius of the round bottom)² (height) .
He starts with a small cone, he then adds clay to it to make it higher.
The question is: How much clay does he ADD to the short one,
to make the bigger one ?
Use the formula to find the volume of the short one.
Use the formula again to find the volume of the bigger one.
Then SUBTRACT the smaller volume from the bigger volume.
THAT's how much clay he has to ADD.
Notice that the new built-up cone has the same radius
but more height than the first cone.
_______________________________________
Don't worry if you don't understand this.
The answer will be this number:
(1/3) (pi) (radius²) (height of the big one minus height of the small one).
Plug in what ever value is the function of g for x and solve:
g(3) = |3-3| = |0| = 0
g(-3) = |(-3)-3| = |(-6)| = 6
Hope this helps :)
You need to use the topic Straight line and find the y-intercept and gradient.
