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3 years ago

Which ordered pair is a solution to this system of equations?y = x2 – 3x + 3 y = 3x – 6

Mathematics

1 answer:

RSB [31]3 years ago

8 0

Assuming the first term in the first equation is actually x^2, then the ordered pair is C. (3, 3).

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PLZ ANSWER ASAP! 15 points

Mariulka [41]

Answer: THe last one.

Step-by-step explanation:

This question is, in essence, basically asking which numbers are less than -3. Looking at the numbers, it is clearly the last set.

5 0

3 years ago

Quadrilateral A and B are scaled copies of each other. Quadrilateral A has side lengths of 2, 3, 5, and 6. Quadrilateral B's lon

harkovskaia [24]

Answer:

Quadrilateral A has side lengths 2, 3, 5, and 6. Quadrilateral B has side lengths 4, 5, 8, and 10. Could one of the quadrilaterals be a scaled copy of the other ...

4 0

3 years ago

What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.

Paul [167]

Answer:

1.60 $ft^{2}$

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = $\pi$ $r^{2}$

So that,

for the inner part of the pipe,

r = $\frac{diameter}{2}$

= $\frac{5}{2}$

= 2.5 feet

Area of the inner part of the pipe = $\pi$ $r^{2}$

= 3.14 x $(2.5)^{2}$

= 19.625

Are of the inner part of the pipe is 19.63 $ft^{2}$ .

Total diameter of the pipe = 5 + 0.2

= 5.2 feet

r = $\frac{5.2}{2}$

= 2.6 feet

Area of the pipe = $\pi$ $r^{2}$

= 3.14 x $(2.6)^{2}$

= 21.2264

Area of the pipe is 21.23 $ft^{2}$ .

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

= 21.2264 - 19.625

= 1.6014

The cross sectional area of the cement pipe is 1.60 $ft^{2}$ .

6 0

2 years ago

Can anybody help with this

iris [78.8K]

100 =d

your welcome lol

8 0

3 years ago

One card is drawn at random from a deck of 52 cards. The first card is not replaced, and a second card is drawn. (Enter your pro

andriy [413]

Answer:

We have a 52 card deck:

a. Find the probability that both cards are clubs.

The probability of drawing a club will be equal to the quotient between the number of clubs in the deck divided by the total number of cards in the deck.

We initially have 13 clubs and 52 cards, then the probability of drawing a club in the first draw is:

p1 = 13/52.

Now the probability of drawing a club in the second draw will be:

p2 = 12/51

Where each number is decreased by one because we already drew a card, and that card was a club.

Then the probability of both events happening will be equal to the product of the individual probabilities:

P = p1*p2 = (13/52)*( 12/51) = 0.059

b. Find the probability that the first card is a spade and the second is a club.

Same as before, the probability of first drawing a spade is:

p1 = 13/52

And the probability of drawing a club in the second draw will be:

p2 = 13/51

This case differs from the prior one because for the second draw we have 13 clubs in the deck, and as we already drew one card (that was not a club) the total number of cards in the deck is 51.

Now the joint probability will be:

P = p1*p2 = (13/52)*(13/51) = 0.064

3 0

3 years ago