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Elenna [48]
3 years ago
12

Can you answer these please! I really need help! And I'm so not even joking only 16 17 18 and 19

Mathematics
1 answer:
frozen [14]3 years ago
4 0
16.) 38.25= 4.25h
divide by 4.25 on both sides
       h=9
17.)214= 1/3x
divide by 1/3 on both sides
x= 642
18.)70= 2x
divide by 2 on both sides
x=35
19.)18.69= 5.25+1.68x
subtract 5.25 on both sides
13.44=1.68x
divide by 1.68 on both sides
x=8
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Write the definition of a function minmax that has five parameters. the first three parameters are integers. the last two are se
Nitella [24]
Hello,


function minmax(int p1,int p2,int p3, int adr_big, int adr_small)
{ int mini=p1,maxi=p1;
 if (p1>p2)   {mini=p2;}
else            {maxi=p2;};

if (p3>maxi) maxi=p3;
if (p3<mini) mini=p3;
*adr_big=maxi;
*adr_small=mini;
};

// main
int a=31,b=5,c=19,big,small;
minmax(a,b,c,&big,&small);





8 0
4 years ago
g natasha is in a class of 30 students that selects 4 leaders. How many ways are there to select the 4 leaders so that natasha i
Papessa [141]

Answer:

<h2>3,654 different ways.</h2>

Step-by-step explanation:

If there are 30 students in a class with natasha in the class and natasha is to select four leaders in the class of which she is already part of the selection, this means there are 3 more leaders needed to be selected among the remaining 29 students (natasha being an exception).

Using the combination formula since we are selecting and combination has to do with selection, If r object are to selected from n pool of objects, this can be done in nCr number of ways.

nCr = n!/(n-r)!r!

Sinca natasha is to select 3 more leaders from the remaining 29students, this can be done in 29C3 number of ways.

29C3 = 29!/(29-3)!3!

29C3 = 29!/(26!)!3!

29C3 = 29*28*27*26!/26!3*2

29C3 = 29*28*27/6

29C3 = 3,654 different ways.

This means that there are 3,654 different ways to select the 4 leaders so that natasha is one of the leaders

5 0
3 years ago
Directions: Answer true or false. If false, provide a counterexample.
Arisa [49]

Natural numbers are closed under division: false.

A set is closed under a certain operation if the results of that operation are always inside that set.

So, if natural numbers were closed under division, the division of two natural numbers would always be a natural number.

You have plenty of counterexamples, to pick one you may divide any odd number by 2: 5/2 is not a natural number.

Negative numbers are closed under addition: true.

Let m,n be two positive numbers. So, -m,-n are two negative numbers. Their sum is

(-m)+(-n)=-(m+n)

And since m+n is positive, we deduce that -(m+n) is negative, so the sum of two negative numbers is still negative.

Prime numbers are closed under subtraction: false.

This would mean that the subtraction of two primes is also a prime. Again, there are many counterexamples: 7 is prime and so is 3, but their difference 7-3 is 4, which is not prime.

7 0
3 years ago
Is the function linear or exponential f(x) =8(1.2)
777dan777 [17]
Linear is the correct answer because you don't see any exponents
7 0
3 years ago
Estimate how many times larger 6.1 x 1
marusya05 [52]
1) 4
2) 2

Pretty sure thats right
7 0
3 years ago
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