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3 years ago

What is the slope of the line perpendicular to the y 2 - 9.

Mathematics

1 answer:

yulyashka [42]3 years ago

6 0

Answer:

Step-by-step explanation:

perpendicular slopes multiply to -1

lets say m is the missing slope

-1/5*m=-1 so m=5

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Suppose 3x+4y=52 and 5x+y= 30. What is the value of 8x-2y

grandymaker [24]

Answer:

Step-by-step explanation: If you made x, 4 and y, 5 then all equations would make sense if you change the letters to numbers.

(3*4)+(4*10)=52

and

(5*4)+10=30

then

6 0

3 years ago

Can someone help me??

xeze [42]

B because plants are often identical to or a clone of their parent and one parent is involved.

7 0

2 years ago

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What is the classifications of polynomials -gh4i+3g5

Step2247 [10]

Answer:

The polynomial -gh⁴i + 3g⁵ is a binomial, since it has two terms

Degree of polynomial: degree of a polynomial is the term with highest of exponent.

Degree of binomial -gh⁴i + 3g⁵ = 6

1st term(-gh⁴i ) = (power of g = 1, power of h = 4, power of i = 1)

2nd term(3g⁵) = (power of g = 5)

the polynomial -gh⁴i + 3g⁵ is a 6 degree binomial.

5 0

3 years ago

Amount of money you earn per hour if you make d dollars in 15 hours??

KATRIN_1 [288]

The answer to the question is d/15

7 0

3 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

Read 2 more answers