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marshall27 [118]
3 years ago
8

Help it doesn't work

Mathematics
1 answer:
kati45 [8]3 years ago
6 0
What are you taking about friend???
You might be interested in
Find the product. (–7t – 5v)(–4t – 3v)
Anuta_ua [19.1K]

Answer: Option B is the correct answer

Step-by-step explanation:

The given expression is

(–7t – 5v)(–4t – 3v).

The product will be a quadratic equation (having 2 as the highest power)

To find the product, we would expand the brackets

(-7t × -4t )+ (-7t × -3v) + (-5v × -4t) + (-5v × - 3v)

= (- -28t^2) +(- -21tv) + (- - 20tv) +(- -15v^2)

Recall, negative × negative equals positive.

=28t^2 +21tv +20tv+ 15v^2)

Collecting like terms, we add all terms containing the same letters together

28t^2 + 41tv + 15v^2

Option B is the correct answer

7 0
3 years ago
Write a unit rate for the situation. 1080 miles on 15 gallons <br> Miles per gallon
MissTica

Answer:

72 miles / gallon

Step-by-step explanation:

1080/15 = x/1

1080 / 15 = 72

72 miles on 1 gallon

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

7 0
3 years ago
Read 2 more answers
The number of patients treated at Dr. Martin's dentist office each day was recorded for nine days. These are the data: 6, 6, 6,
zmey [24]
Mean = 5.555555555.....
median = 6
mode = 6
range= 1
3 0
4 years ago
If you draw four cards at random from a standard deck of 52 cards, what is the probability that all 4 cards have distinct charac
mezya [45]

There are \binom{52}4 ways of drawing a 4-card hand, where

\dbinom nk = \dfrac{n!}{k!(n-k)!}

is the so-called binomial coefficient.

There are 13 different card values, of which we want the hand to represent 4 values, so there are \binom{13}4 ways of meeting this requirement.

For each card value, there are 4 choices of suit, of which we only pick 1, so there are \binom41 ways of picking a card of any given value. We draw 4 cards from the deck, so there are \binom41^4 possible hands in which each card has a different value.

Then there are \binom{13}4 \binom41^4 total hands in which all 4 cards have distinct values, and the probability of drawing such a hand is

\dfrac{\dbinom{13}4 \dbinom41^4}{\dbinom{52}4} = \boxed{\dfrac{2816}{4165}} \approx 0.6761

4 0
2 years ago
Which values of a and b make this system of equations have infinitely many solutions?
Afina-wow [57]

9514 1404 393

Answer:

  a = 3, b = -8

Step-by-step explanation:

Solving the first equation for y, we get ...

  2y +16 = 6x . . . . . given

  y = 8 +3x . . . . . . . divide by 2

  y = 3x -8 . . . . . . . subtract 8

In order for the system of equations to have infinitely many solutions, the second equation must be the same as this:

  y = ax +b

  a = 3, b = -8

6 0
3 years ago
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