I WILL MARK BRAINIEST

aniked [119]

FIRST QUESTION
Given points are:
A(2, -4) and B(6, 2)
Now, Use the distance formula.
distance formula = $\sqrt{ (x_{2}- x_{1})^{2} + ( y_{2} - y_{1} )^{2} }$

Now, plug the values into the formula, So,
distance = $\sqrt{ (6- 2)^{2} + ( 2 - (-4))^{2} }$

= $\sqrt{ (6- 2)^{2} + ( 2 +4))^{2} }$

= $\sqrt{ (4)^{2} + ( 6))^{2} }$

= $\sqrt{ 16+36}$

= $\sqrt{52}$

= $2 \sqrt{13}$

So, the length of AB is $2 \sqrt{13}$ .

THIRD QUESTION
Two points given are:
A(3, -2) and B(1, 1)
Also given that B is the midpoint of AC.

Let, the co-ordinates of C be C(a, b).
Now, using midpoint formula,
Midpoint = $(\frac{ x_{1}+ x_{2} }{2} , \frac{ y_{1}+ y_{2} }{2} )$

$(1, 1)=(\frac{ 3+ a }{2} , \frac{ -2+b }{2} )$

Now, equaling the ordered pair, we have,

$1=\frac{ 3+ a }{2}$ .............equation (1)

$1=\frac{ -2+b }{2}$ ................equation (2)

Now, taking equation (1)
$1=\frac{ 3+ a }{2}$

$1*2=3+a$

$2-3=a$

$a=-1$

Now, taking equation (2)
$1=\frac{ -2+b }{2}$

$1*2=-2+b$

$2+2=b$

$b=4$

So, the co ordinates of C are (a, b) which is (-1 , 4)


SECOND QUESTION:
Given equations are:
2x + 3y = 14.....................equation (1)
-4x + 2y = 4 .....................equation (2)
Taking equation (2)
-4x + 2y = 4
2y = 4 + 4x
y = (4 + 4x) / 2
y = 2 + 2x .......................equation (3)
Now, Taking equation (1)
2x + 3y = 14
Substituting the value of y from equation (3), we get,
2x + 3(2 + 2x) = 14
2x + 6 + 6x = 14
8x = 14 - 6
x = (14 - 6) / 8
x = 1

Taking equation (3)
y = 2 + 2x
Now, substituting the value of x in equation (3), we get,
y= 2 + 2 (1)
y = 2 + 2
y = 4

So, x=1 and y=4

6 0

3 years ago

Alexandra has some dimes and some quarters. She has no more than 21 coins worth a minimum of $3.75 combined. If Alexandra has 4

olga55 [171]

Answer:

A minimum of 10 dimes and 11 quarters is what Alexandra will have

Step-by-step explanation:

Let

d = number of dimes

q = number of quarters

Since she has 21 coins altogether,

d + q = 21------------------------equation 1

If these coins are worth $3.75 then

0.10 x d + 0.25 x q = 3.75

which is 0.10d +.25q =3.75 --------------------------equation 2

where $.10 is the value of one dime and $.25 is the value of one quarter

make d the subject of formula from equation 1 d = 21 -q----------equation 3

insert it in equation 2

0.10d +0.25q =3.75

0.10(21-q) + 0.25q = 3.75

0.1(21)-0.1q+0.25q=3.75

2.1 +0.15q = 3.75

0.15q = 3.75-2.1 = 1.65

q = 1.65/0.15 =165/15 =11

since we have the value of q insert in equation 3

d = 21 - q

d = 21-11

d = 10

Alexandra has 10 dimes and 11 quarters.

from my calculation i can see that the a minimum of 10 dimes and 11 quarters is what Alexandra will have

4 0

3 years ago

A 30-year-old female purchased a 20-Year Endowment insurance policy at the age of 23. The face value of the policy was $78,250.

FrozenT [24]

Answer:

B

Step-by-step explanation:

Annual Premium of $2,957.85; 26 years and 10 days

8 0

3 years ago

Read 2 more answers

Using the given points and line, determine the slope of the line.

nekit [7.7K]

M = y2-y1 / x2-x1
= 2-1 / 1-2
= 1 / -1
= -1

7 0

3 years ago

Read 2 more answers

Naomi invested $3,425 in an account that pays 3% simple annual interest. what was the total balance in the account after 15 year

Alenkinab [10]

Each year, your $3,425 which earn 3% simple interest earns $102.75. Multiplying this value by 15 gives $1541.25. Then, adding the latest value we have with the initial invest will finally give us $4966.25. Thus, after 15 years Naomi will have $4966.25.

6 0

3 years ago

Read 2 more answers