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Triss [41]
2 years ago
6

Please help me this is my last question!!​

Mathematics
1 answer:
grin007 [14]2 years ago
3 0

Answer:

You can solve it by saying half times base times perpendicular high

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If p q r is prime numbers such that pq+r=73, what is the least possible value of p+q+r
Jobisdone [24]
The answer to this question would be: p+q+r = 2 + 17 + 39= 58

In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
Since 73 is an odd number, it is clear that one of the p q r needs to be an even number. 

There is only one odd prime number which is 2. If you put 2 in the r the result would be:
pq+2= 73
pq= 71
There will be no solution for pq since 71 is prime number. That mean 2 must be either p or q. Let say that 2 is p, then the equation would be: 2q + r= 73

The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
r= 73-2q
r= 73- 2(17)
r= 73-34=39

p+q+r = 2 + 17 + 39= 58
6 0
3 years ago
Read 2 more answers
What is this?<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bf%7D%7B4%7D%20%20-%205%20%3D%20%20-%209" id="TexFormula1" title=
tigry1 [53]

Hey there!

<h2>ANSWER: f=-16</h2><h2>EXPLANATION:</h2>

\frac{f}{4}-5=-9

Simplify both sides and you get:

\frac{1}{4}f-5=9

Now add 5 to both sides and you get:

\frac{1}{4}f=-4

Now you have to multiply both sides and you get:

f=-16(ANSWER)

Hope this helps!

\text {-TestedHyperr}

6 0
3 years ago
Pls help i will give the brielliast answer
marishachu [46]
The answer is the last one the one that starts with king’s speech
6 0
3 years ago
a product code consists of picking a two-digit number (including 00) and 3 not necessarily distinct letters. How many codes are
Brrunno [24]

Answer:

The codes are:

N,_,_,_

Where each _ represents a possible letter, out of 26.

And N is a number of two digits.

So in total we can think that we have 5 empty slots.

_,_,_,_,_

In the first slot we can put any digit between 0 and 9, so here we have 10 options.

In the second slot we can put any digit between 0 and 9, so here we have 10 options.

In the third slot we can put any letter, so here we have 26 options (and exactly the same for the fourth and fifth slots)

The total number of different combinations is equal to the product of the number of options for each slot.

C = 10*10*26*26*26 = 1,757,600

Now, if we want to have at least one nine, we can fix it in the first slot.

Then we have:

in the first slot we have only one option (the 9)

In the second slot we can put any digit between 0 and 9, so here we have 10 options.

In the third slot we can put any letter, so here we have 26 options (and exactly the same for the fourth and fifth slots)

The number of combinations is:

C = 1*10*26*26*26 = 175,760

But we also should consider the case where we fix the 9 in the second slot, so the actual number of combinations is twice the number above.

C = 2*175,760 = 351,520

The probability that the code does not contain the number 9.

Now in the first slot we have only 9 options, all the whole numbers between 0 and 8.

The same for the second slot, 9 options.

For the third, fourth and fifth slot is the same as before.

The total number of combinations is:

C = 9*9*26*26*26 = 1,423,656

4 0
3 years ago
I really need to understand this question! Please answer
mars1129 [50]

Answer:

- 3

Step-by-step explanation:

Using the law of logarithms

log_{b} x = n ⇔ x = b^{n}

let log_{b}( \frac{1}{b^3}) = n

log_{b}( b^{-3}) = n, then

b^{-3} = b^{n}

Since bases are the same on both sides then equate the exponents

⇒ n = - 3 → (4)

4 0
3 years ago
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