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3 years ago

Which point lies on the line

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

7 0

Y + 8 = 4(x - 5)

distribute the 4 to both x and -5

4(x) = 4x
4(-5) = -20

y + 8 = 4x - 20

isolate the y, subtract 8 from both sides

y + 8 (-8) = 4x - 20 (-8)

y = 4x - 28

plug in the x given, if the y matches the answer given, it is the correct answer.

-8 = 4(5) - 28

-8 = 20 - 28

-8 = -8 (True)

D) (5,-8) should be your answer

hope this helps

zhannawk [14.2K]3 years ago

4 0

The point (5,-8) or D lies on the line

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4 years ago

Anna drives 15,500 miles a year for work. On average, how many miles does she drive a week?

I am Lyosha [343]

Answer:

206 miles

Step-by-step explanation:

People generally work 5x a week so.. (15000 / 52)

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3 years ago

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Answer:

0.22m

Step-by-step explanation:

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2 years ago

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About 19% of the population of a large country is hopelessly romantic. If two people are randomly selected, what is the probab

rjkz [21]

Answer:

The probability that both will be hopelessly romantic is

P(X = 2) = 0.0361

b)

The probability that at least one person is hopelessly romantic is

P( X>1) = 0.3439

Step-by-step explanation:

Given data population proportion 'p' = 19% =0.19

q = 1-p = 1- 0.19 =0.81

Given two people are randomly selected

Given n = 2

Let 'X' be the random variable in binomial distribution

$P(X=r) =n_{C_{r} } p^{r} q^{n-r}$

The probability that both will be hopelessly romantic is

$P(X= 2) =2_{C_{2} } (0.19)^{2} (0.81)^{2-2}$

P(X = 2) = 1 × 0.0361

The probability that both will be hopelessly romantic is

P(X = 2) = 0.0361

b)

The probability that at least one person is hopelessly romantic is

P( X>1) = 1-P(x<1)

= 1 - ( p(x =0)

= $1- 2_{C_{0} } (0.19)^{0} (0.81)^{2-0}$

= 1 - (0.81)²

= 1 -0.6561

= 0.3439

The probability that at least one person is hopelessly romantic is

P( X>1) = 0.3439

7 0

3 years ago

Could someone help me, please

SashulF [63]

Answer:

Q1: No solutions

Q2: x=2, y=-2

Q3: x=1, y=4

Step-by-step explanation:

Q1: Found out 0y = 18 which is impossible therefore no solution.

Q2:

In the first equation: $2x - 4y = 12$

Let's express x in terms of y by changing the subject. Follow my steps.

$2x = 12+4y$

$x = \frac{12+4y}{2}$

Now let's put the x above into the second equation: $-14x-4y=-20$

It becomes: $-14(\frac{12+4y}{2})-4y=-20$ . Follow my steps.

$-7(12+4y)=-20+4y$

$-84-28y=-20+4y$

$-32y=64$

$y=-2$

Put y = -2 into the first equation:

$2x-4(-2)=12$

$2x+8=12$

$2x=12-8$

$2x=4$

$x=2$

Q3:

Multiply the first equation by 3 and the second equation by 7, so that both are 21x.

1st equation:
$3(7x-3y)=3(-5)$
$21x-9y=-15$

2nd equation:

$7(3x+2y) = 7(11)$

$21x+14y = 77$

We can now substract the 2nd equation with the 1st equation so that both 21x cancels out and only the variable y remains.

$21x-9y-(21x+14y)=-15-77$

$21x-9y-21x-14y=-92$

$-23y=-92$

$y=4$

Put y = 4 into the second equation to find x.

$3x+2(4)=11$

$3x+8=11$

$3x=11-8$

$3x = 3$

$x=1$

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3 years ago