If you look at the model RM is the same length as PM. PM=10 so RM=10.
Answer:
In quadrilateral ABCD we have
AC = AD
and AB being the bisector of ∠A.
Now, in ΔABC and ΔABD,
AC = AD
[Given]
AB = AB
[Common]
∠CAB = ∠DAB [∴ AB bisects ∠CAD]
∴ Using SAS criteria, we have
ΔABC ≌ ΔABD.
∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.
∴ BC = BD.
Answer:
Price of widget to break even= $37.9
Step-by-step explanation:
We are told that the equation representing the amount of profit, y, made by the company, in relation to the selling price of each widget, x is;
y = -8x² + 348x - 1705
Now, the company will break even when it has made no profit. That is when, y = 0
Thus;
0 = -8x² + 348x - 1705
Rearranging,
8x² - 348x + 1705 = 0
Using quadratic formula ;
x = [-b ± √(b² - 4ac)]/2a
x = [-8 ± √(-348² - 4•1•1705)]/(2 x 8)
x = $5.63 or $37.87
We'll use $37.87 because it is the highest price for which no profit is made, and higher price means that we could sell least number of products to earn a certain amount of money.
We are told to approximate to nearest cent. Thus,
Price of widget = $37.87 ≈ $37.9
<u>Unit rate 4:</u>
The table
d n
4 1
8 2
16 4
<u>Unit rate 1/4:</u>
Table
d n
1 4
4 16
16 64
and
The equation n=4d
<u>Unit rate 16:</u>
Equation:
d = 16n
Step-by-step explanation:
The unit in the unit rate is dollars per unit ounces which means that the unit rate calculated by
So,
The first option is the equation:
The unit rate is 16 dollars per ounce
The second option is the table:
d n
1 4
4 16
16 64
We can take any pair of values of d and n from the table to calculate the unit rate
So taking
d = 1
n=4
The unit rate for table is: 1/4 dollars per ounces
Third option is the equation:
n = 4d
dividing both sides by n
dividing both sides by 4
the unit rate is 1/4 dollars per ounce
Fourth option is the table:
d n
4 1
8 2
16 4
We will take any pair of d and n to find the unit rate
So,
Taking
d = 4
n =1
The unit rate is 4 dollars per unit ounce.
Keywords: Unit rate, units
Learn more about unit rate at:
#LearnwithBrainly
Team A) 45 people
Team B) 55 people
A)There are two ways to solve this problem, finding the number of combinations possible for Team B, or the number of combinations possible for Team A.
Team A
It's a given that 20 mathematicians are on team A, which leavs the other 25 people for team A to be chosen from a pool of 80 (100- 20 mathletes)
80-C-25 = 80! / (25!/(80-25)!) =<span>363,413,731,121,503,794,368
</span>or 3.63 x 10^20
Solving using Team B
Same concept, but choosing 55 from a pool of 80 (mathletes excluded)
80-C-25 = 80! / (55!(80-55!) = 363,413,731,121,503,794,368
or 3.63 x 10^20
As you can, we get the same answer for both.
B)
If none of the mathematicians are on team A, then we exclude the 20 and choose 45:
80-C-45 = 80! / (45!(80-45)!) = <span>5,790,061,984,745,3606,481,440
or 5.79 x 10^22
Note that, if you solve from the perspective of Team B (80-C-35), you get the same answer</span>