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deff fn [24]
3 years ago
12

Will make brainliest if answered correctly

Mathematics
2 answers:
slava [35]3 years ago
6 0
The average speed was 62.5 kilometers.
stealth61 [152]3 years ago
3 0

Answer:

3.75Km

Step-by-step explanation:

to find Average speed you do distance over time so it would be 225/25=3.75

You might be interested in
If sin 0(theta)=2/3 and tan 0(theta)<0. What is the value of cos 0(theta)
BigorU [14]

Answer:

cos Ф = adj / hyp = √5 / 3

Step-by-step explanation:

If sin Ф is +, then Ф must be in either Quadrant 1 or Quadrant 2.

If tan Ф < 0, then Ф must be in either Quadrant 2 or Quadrant 3.  

So we conclude that Ф is in Quadrant 2.

If sin Ф = opp / hyp = 2/3, then opp = 2 and hyp = 3, and adj is found using the Pythagorean Theorem:

adj = √( 3² - 2² ) = √( 5 )

With adj = √5 and hyp = 3, cos Ф = adj / hyp = √5 / 3

7 0
2 years ago
(-3x)^2<br><br>wbat is this please ​
valentina_108 [34]

if you want me to solve (-3x)²

the answer is 9x²

3 0
3 years ago
Read 2 more answers
A farming council takes a large random survey of US adults in which they ask how many times they ate a
Lady_Fox [76]

Answer:

Histogram A

Step-by-step explanation:

Because I was trying to find the answer but I got it wrong and it told me the answer <3

6 0
3 years ago
eight cards are drawn from a standard deck of 52 cards. how many hands off with cards contain exactly three queens and three jac
d1i1m1o1n [39]

Answer:

15,136 hands off with cards contain exactly three queens and three jacks.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Standard deck:

4 queens and 4 jacks.

The other 52 - 8 = 44 cards are neither queens nor jacks.

Wow many hands off with cards contain exactly three queens and three jacks?

3 queens from a set of 4.

3 jacks from a set of 4.

2 other cards(not queens neither jacks) from the other 44. So

C_{4,3}C_{4,3}C_{44,2} = \frac{4!}{1!3!} \times \frac{4!}{1!3!} \frac{44!}{2!42!} = 4*4*22*43 = 15136

15,136 hands off with cards contain exactly three queens and three jacks.

4 0
3 years ago
Can someone help me with E please I don’t know what operation to use
gtnhenbr [62]
The first question asks who sold more rolls. So start with figuring out how many Christie sold.
5 total - 1 2/3 left = 3 1/3 sold
you can convert the numbers to improper fractions with the same denominator. Like this:
5 x (3/3) - (3+2)/3
15/3 - 5/3 = 10/3
10/3 = 3 1/3

So now we know Christie sold more because 3 1/3 dozen is more than 2 1/2 dozen.

The part asks how many more.. Subtract the amounts the two girls sold.

3 1/3 - 2 1/2
10/3 x (2/2) - 5/2 x (3/3)
20/6 - 15/6 = 5/6

Christie sold 5/6 dozen more rolls. A dozen is 12 rolls so if you wanted to go further you just multiply 12 x 5/6 = 10 rolls
4 0
3 years ago
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