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Gelneren [198K]
2 years ago
12

11. S(-4,-6) and T(-7, -3); Find R. What is the missing endpoint if S is the midpoint of RT

Mathematics
1 answer:
Nezavi [6.7K]2 years ago
5 0

answer is (-7,-9)

step by step:-

S is the midpoint of RT so ,

s=\frac{R+T}{2 }

-4=\frac{x-7}{2}  ,     -6=\frac{y-3}{2}

-14=x-7  ,   -12=y-3

x=-7          ,    y=-9

R= (-7,-9)

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If 49x2 + 28x - 10 is rewritten as p2 + 4p -10, what is p in terms of<br> p=
madam [21]

Answer:

p=7x

Step-by-step explanation:

49x^[2] + 28x - 10 = p^[2] + 4p -10

This equation is in the form a^[2]x + bx + c.

<u><em>The 'c' is common for both equations, this means the 'a' and 'b' must also be common. </em></u>

There are two ways to find p: 'a' or 'b'

<u>a method</u>

49x^[2] = p^[2]

=> The square root of both sides = 7x = p

<u>b method</u>

28x = 4p

28x/4 = 4p/4

7x = p

6 0
1 year ago
A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d
tankabanditka [31]
The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
thus, the sample space of the experiment is 286

A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is 

</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls. 

</span>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<span>
the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in 

</span>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
6 0
2 years ago
Read 2 more answers
The table shows expressions to represent the number of students involved in different activities. The number of students involve
Maurinko [17]

Answer:

6n + 7 = 7n

Step-by-step explanation:

Number of students involved in sports = 5n + 7

Number of students involved in student council = n

Number of students involved in sports and student council = (5n + 7) + n = 5n + 7 + n = 6n + 7

Number of students involved in band = 3n - 2

Number of students involved in drama club = 2(2n + 1)

Number of students involved in band and drama club = (3n - 2) + 2(2n + 1) = 3n - 2 + 4n + 2 = 7n

Equation to model number of students involved in sports and student council which equal number of students involved in band and drama club:

6n + 7 = 7n

6 0
2 years ago
If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle,
liubo4ka [24]
Given:
Segment AC = 130 feet
Segment CD = 70 feet

I think that I'll be using the Pythagorean Theorem in finding the value of r. r will be the hypotenuse

Segment CE = (r - 70 feet)

r² = a² + b²
r² = 130² + (r-70)²
r² = 16,900 + (r-70)(r-70)
r² = 16,900 + r² - 70r - 70r + 4900
r² - r² + 140r = 16,900 + 4,900
140r = 21,800
r = 21,800/140
r = 155.71 feet

The radius of the circle is 155.71 feet.

8 0
2 years ago
Read 2 more answers
A collection of 77 quarters and dimes is worth $12.50. How many quarters and dimes are there? Set it up plz
mihalych1998 [28]
So let's say quarters = x and dimes = y. You'd then put x + y = 77 (for a total of 77 coins). And your second equation would be .25x + .10y = 12.50 (.25 for the total value a quarter holds and .10 for a total value a dime holds). Then you'd do matrices on a calculator.
4 0
2 years ago
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