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3 years ago

X+y=4x²+y²=40solve simulation

Mathematics

1 answer:

nikdorinn [45]3 years ago

7 0

Use the website math papa it’ll give you the answers for that I promise

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F(x) = 7 - 5x + 7x2<br> find f(-9)

Law Incorporation [45]

Answer:

Step-by-step explanation:

F(x) = 7 - 5x + 7x²

f(-9) = 7 - 5*(-9) + 7*(-9)²

= 7 + 45 + 7*81

= 7 + 45 + 567

= 619

7 0

3 years ago

Read 2 more answers

X^2+y^2-4y-15=6

myrzilka [38]

Answer:

x^2 + y^2 - 4y - 15 = 6

x^2 + y^2 - 4y = 21

x^2 + y^2 - 4y + 4 = 25

x^2 + (y - 2)^2 = 5^2

r = 5

8 0

2 years ago

In an algebra class last year, students who earned grades of A spent 4.7 hours more on homework per week than students who earne

Neko [114]

<span>A = 4.7 + C
where A represents the students who got A's, and C is the number of hours that the C-grade students spent on homework per week</span>

8 0

2 years ago

Can some please help me

qaws [65]

Answer:

i think it is D

3 0

2 years ago

An envelope contains three cards: a black card that is black on both sides, a white card that is white on both sides, and a mixe

Over [174]

Answer:

There is a 2/3 probability that the other side is also black.

Step-by-step explanation:

Here let B1: Event of picking a card that has a black side

B2: Event of picking a card that has BOTH black side.

Now, by the CONDITIONAL PROBABILITY:

$P(B_2/B_1 ) = \frac{P(B_1\cap B_2)}{P(B_1)}$

Now, as EXACTLY ONE CARD has both sides BLACK in three cards.

⇒ P (B1 ∩ B2) = 1 /3

Also, Out if total 6 sides of cards, 3 are BLACK from one side.

⇒ P (B1 ) = 3 /6 = 1/2

Putting these values in the formula, we get:

$P(B_2/B_1 ) = \frac{P(B_1\cap B_2)}{P(B_1)} = \frac{1}{3} \times\frac{2}{1} = \frac{2}{3}$

⇒ P (B2 / B1) = 2/3

Hence, there is a 2/3 probability that the other side is also black.

5 0

3 years ago

X+y=4x²+y²=40solve simulation​

X+y=4x²+y²=40solve simulation