Answer:
Please refer to https://www.algebra.com/algebra/homework/equations/Equations.faq.question.1134615.html
Answer:
ST = 8 m
Step-by-step explanation:
Since ∆PQR is congruent to ∆STU, therefore, the three sides of ∆PQR would be congruent or equal to the corresponding three sides of ∆STU.
PQ corresponds to ST.
Therefore, PQ is congruent to corresponding side ST, which is 8 m
ST = 8 m
If parent functin is f(x)=|x|
it is moved to the left 2 units
vertically streched by a factor of 3
and moved up by 4 units in that order
because
to move a function to left c units, add c to every x
to vertically strech function by factor of c, multiply whole function by c
to move funciotn up c units, add c to whole function
so it is 2 to the left, verteically streched by a factor of 3 then moved up 4 units
Answer:
150m^2
Step-by-step explanation:
Given data
a=8m
b=17m
h=12m
We know that the expression for the area of a trapezoid is given as
Area= (a+b/2)*h
substitute
Area= (8+17/2)*12
Area= 25/2 *12
Area= 12.5*12
Area= 150 m^2
Hence the area is 150m^2
Answer:
<h2>35 different ways</h2>
Step-by-step explanation:
Since there are 7 students in a classroom to fill a front row containing 3 seats, we will apply the combination rule since we are to select 3 students from the total number of 7 students in the class.
In combination,<em> if r objects are to be selected from a pool of n objects, this can be done in nCr number of ways.</em>
<em>nCr = n!/(n-r!)r!</em>
Selecting 3 students from 7 students to fill the seats can therefore be done in 7C3 number of ways.
7C3 = 7!/(7-3)!3!
7C3 = 7!/(4)!3!
7C3 = 7*6*5*4!/4!*3*2
7C3 = 7*6*5/6
7C3 = 7*5
7C3 = 35
<em>Hence there are 35 different ways that the student can sit in the front assuming there are no empty seats.</em>
