Answer:
Median = 7
upper quartile = 10
lower quartile is 6
Step-by-step explanation:
The median is the middle line in the box
Median = 7
The upper quartile is the right line of the box
upper quartile = 10
The lower quartile is the left line of the box
lower quartile is 6
A. Factor 28 + 7g 7(g <span>+ 4)
add
28 </span><span>+ 7g
= 7 (g </span>+ 4)
B. (12r + 15) – 65 Remove parentheses
12r+15-65
= 12r - 50
C. 24 – (8p – 12)
24 - (8p - 12) = -8p + 36
24 - (8p - 12) -(8p - 12) 8p + 12
= 24-8p+12
Simplify 24-8p+12
Group like terms
-8p+12+24
Add the numbers 12+24=36
Answer is -8p+36
Answer:
x = ± 10
Step-by-step explanation:
Given
x² - 100 = 0 ( add 100 to both sides )
x² = 100 ( take the square root of both sides )
x = ±
← note plus or minus, hence
x = ± 10
Answer:
x=66⁰
Step-by-step explanation:
they are co interior angles,
180-114=66⁰
To check,
114+66=180
brainlist me?
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
__
You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
__
Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.