Answer:


Science class had a higher average
Step-by-step explanation:
Given

g(x) table
Solving (a): f(2)
We have:

So:



Solving (b): g(2)
From the given table.

Solving (c): f(4) or g(4); which is greater

So:



For g(4): Notice that in the table of g(x); g(x) increases by 2 when x increases by 1
This means that:

So




<em>Hence, g(4) i.e. Science class is greater</em>
Answer:
Step-by-step explanation:
Since none of the answer choices match the drawing of the gardener, we assume the question is referring to the drawing of the partner.
The gardener's drawing is 1/4 of actual size. So, in terms of the gardener's drawing, actual size is ...
gardener's drawing = (1/4)actual size
actual size = 4(gardener's drawing)
__
The partner's drawing is 1/20 of actual size, so is ...
partner's drawing = actual size/20 = (4(gardener's drawing))/20
partner's drawing = (4/20)(gardener's drawing)
partner's drawing = (gardener's drawing)/5
__
Then the {length, width} of the partner's drawing are ...
partner's drawing {length, width} = {15 in, 10 in}/5 = {3 in, 2 in}
The partner's drawing has a length of 3 inches and a width of 2 inches.
Answer:
There are 25 students in the class.
Step-by-step explanation:
Let x be the number of students in the class
12% failed and the number of students is 3
x*12% = 3
x*.12 = 3
Divide each side by .12
x * .12 /.12 = 3/.12
x =25
There are 25 students in the class
Answer:
x = 14
Step-by-step explanation:
Assume your diagram is like the one below.
The intersecting secant angles theorem states, "When two secants intersect outside a circle, the measure of the angle formed is one-half the difference between the far and the near arcs."
For your diagram, that means
![\begin{array}{rcl}m\angle L &=&\dfrac{1}{2} \left(m \widehat {JM} - m\widehat {PQ}\right)\\\\(3x + 13)^{\circ}& = &\dfrac{1}{2} \left[(8x + 48)^{\circ} - (5x - 20)^{\circ}\right]\\\\3x + 13& = &\dfrac{1}{2}(8x + 48 - 5x + 20)\\\\3x + 13& = &\dfrac{1}{2}(3x + 68)\\\\6x + 26 & = & 3x + 68\\6x & = & 3x + 42\\3x & = & 42\\x & = & \mathbf{14}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7Dm%5Cangle%20L%20%26%3D%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%28m%20%5Cwidehat%20%7BJM%7D%20-%20m%5Cwidehat%20%7BPQ%7D%5Cright%29%5C%5C%5C%5C%283x%20%2B%2013%29%5E%7B%5Ccirc%7D%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%5B%288x%20%2B%2048%29%5E%7B%5Ccirc%7D%20-%20%285x%20-%2020%29%5E%7B%5Ccirc%7D%5Cright%5D%5C%5C%5C%5C3x%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%288x%20%2B%2048%20-%205x%20%2B%2020%29%5C%5C%5C%5C3x%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%283x%20%2B%2068%29%5C%5C%5C%5C6x%20%2B%2026%20%26%20%3D%20%26%203x%20%2B%2068%5C%5C6x%20%26%20%3D%20%26%203x%20%2B%2042%5C%5C3x%20%26%20%3D%20%26%2042%5C%5Cx%20%26%20%3D%20%26%20%5Cmathbf%7B14%7D%5C%5C%5Cend%7Barray%7D)
Check:
![\begin{array}{rcl}(3\times14 + 13) & = &\dfrac{1}{2} \left[(8\times14 + 48)^{\circ} - (5\times14 - 20)^{\circ}\right]\\\\42 + 13& = &\dfrac{1}{2}(112 + 48 - 70 + 20)\\\\55& = &\dfrac{1}{2}(110)\\\\55 & = & 55\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%283%5Ctimes14%20%2B%2013%29%20%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%5B%288%5Ctimes14%20%2B%2048%29%5E%7B%5Ccirc%7D%20-%20%285%5Ctimes14%20-%2020%29%5E%7B%5Ccirc%7D%5Cright%5D%5C%5C%5C%5C42%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%28112%20%2B%2048%20-%2070%20%2B%2020%29%5C%5C%5C%5C55%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%28110%29%5C%5C%5C%5C55%20%26%20%3D%20%26%2055%5C%5C%5Cend%7Barray%7D)
It checks.
I think it’s D because the ratio