1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
s344n2d4d5 [400]
3 years ago
15

Liliana used 4 dark power crystals to raise 14 zombie soldiers. She wants to know how many zombie soldiers (z) she can raise wit

h 10 dark power crystals. How many zombie soldiers can Liliana raise with 10 power crystals?
Mathematics
2 answers:
madam [21]3 years ago
8 0
4c = 14z.
1c = 3.5z.
10c = 35z.
So she can raise 35 with 10 power crystals. Hope it helps !:)
andre [41]3 years ago
3 0

Answer:- Liliana can raise 35 zombie soldiers with 10 power crystals.


Explanation:-

Let the number of zombie soldiers raised by 10 dark power crystals be 'x'.

Zombie raised by 4 power crystals = 14

Since the number of zombie and power crystals are in direct proportion.

Therefore, we get

\frac{x}{10}=\frac{14}{4}

Multiply 10 on both the sides, we get

\Rightarrow\ x=\frac{14}{4}\times10=35

Liliana can raise 35 zombie soldiers with 10 power crystals.

You might be interested in
Hi guys<br>I need 3 minus and plus operations with answer thx!​
nata0808 [166]

1999+1001=3000 1091+1009=3000 2000+1000=3000

3000-1001=1999 3000-1009=1091 3000-2000=1000

8 0
2 years ago
PLEASE I NEED HELP! I want to understand this question.
jasenka [17]

Answer:

TP = 5. greater than

Step-by-step explanation:

1. Out of the 4 options, HH, HT, TH, TT, there is one option we want (HT) which is one our of the four options. that means that the theoretical probability is 1/4 = 25%. Since there were 20 flips, the theoretical probability is 25% of 20 which is 5.

2. For experimental probability, its what actually happenned. Out of the 20 flips, 6 were HT so comapared to the Theoretical probability, the experimental probability was higher.

7 0
2 years ago
Let $s$ be a subset of $\{1, 2, 3, \dots, 100\}$, containing $50$ elements. how many such sets have the property that every pair
Tamiku [17]

Let A be the set {1, 2, 3, 4, 5, ...., 99, 100}.

The set of Odd numbers O = {1, 3, 5, 7, ...97, 99}, among these the odd primes are :

P={3, 5, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

we can count that n(O)=50 and n(P)=24.

 

 

Any prime number has a common factor >1 with only multiples of itself.

For example 41 has a common multiple >1 with 41*2=82, 41*3=123, which is out of the list and so on...

For example consider the prime 13, it has common multiples >1 with 26, 39, 52, 65, 78, 91, and 104... which is out of the list.

Similarly, for the smallest odd prime, 3, we see that we are soon out of the list:

3, 3*2=6, 3*3=9, ......3*33=99, 3*34=102.. 

we cannot include any non-multiple of 3 in a list containing 3. We cannot include for example 5, as the greatest common factor of 3 and 5 is 1.

This means that none of the odd numbers can be contained in the described subsets.

 

 

Now consider the remaining 26 odd numbers:

{1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

which can be written in terms of their prime factors as:

{1, 3*3, 3*5, 3*7, 5*5,3*3*3, 3*11,5*7, 3*13, 2*2*3*3, 7*7, 3*17, 5*11 , 3*19,3*21, 5*13, 3*23,3*5*5, 7*11, 3*3*3*3, 5*17, 3*29, 7*13, 3*31, 5*19, 3*3*11}

 

1 certainly cannot be in the sets, as its common factor with any of the other numbers is 1.

3*3 has 3 as its least factor (except 1), so numbers with common factors greater than 1, must be multiples of 3. We already tried and found out that there cannot be produced enough such numbers within the set { 1, 2, 3, ...}

 

3*5: numbers with common factors >1, with 3*5 must be 

either multiples of 3: 3, 3*2, 3*3, ...3*33 (32 of them)

either multiples of 5: 5, 5*2, ...5*20 (19 of them)

or of both : 15, 15*2, 15*3, 15*4, 15*5, 15*6 (6 of them)

 

we may ask "why not add the multiples of 3 and of 5", we have 32+19=51, which seems to work.

The reason is that some of these 32 and 19 are common, so we do not have 51, and more important, some of these numbers do not have a common factor >1:

for example: 3*33 and 5*20

so the largest number we can get is to count the multiples of the smallest factor, which is 3 in our case.

 

By this reasoning, it is clear that we cannot construct a set of 50 elements from {1, 2, 3, ....}  containing any of the above odd numbers, such that the common factor of any 2 elements of this set is >1.

 

What is left, is the very first (and only) obvious set: {2, 4, 6, 8, ...., 48, 50}

 

<span>Answer: only 1: the set {2, 4, 6, …100}</span>

8 0
3 years ago
Solve<br><br> Please help don’t understand
tigry1 [53]

Answer:x=30*

Step-by-step explanation:

90*+30*=120*

*All triangles’ angles = 180*

180*-120*=60*

60*/2=30*

8 0
3 years ago
Crossword Puzzle
Vikki [24]

Answer:

Please read the answers below that solve the crossword puzzle about linear equations. Tested and proved offline!

Step-by-step explanation:

Across:

2. V a r i a b l e

9. S o l v e

10. S u b t r a c t i o n p r o p e r t y (Two words)

Down:

1. M u l t i p l i c a t i o n p r o p e r t y (Two words)

3. A d d i t i o n p r o p e r t y (Two words)

4. L i n e a r

5. D i v i s i o n p r o p e r t y (Two words)

6. L i n e a r i n e q u a l i t y (Two words)

7. C o n s t a n t t e r m (Two words)

8. C o e f f i c i e n t

3 0
3 years ago
Read 2 more answers
Other questions:
  • Diana uses 30 grams of coffee beans to make 48 fluid ounces of coffee. When company comes she makes 100 fluid ounces of coffee.
    9·1 answer
  • A man tied a rope to the top of a tree, which is 'x' m tall. The other end of the rope was tied to
    7·2 answers
  • Find the values of a, b and c.
    12·1 answer
  • Event A: Flipping heads on a coin.Event B: Spinning orange on the spinner shown (4 parts to the spinner) .What is P(A and B)?
    15·1 answer
  • Solve the inequality. Using a verbal statement, in simplest terms, describe the solution of the inequality. Be sure to include t
    5·1 answer
  • Suppose the graph of y= x^2 translated left 2 units, and up 8 units
    14·1 answer
  • the difference of an acute angle and an obtuse angle is an obtuse angle. sometimes, always, or never true.​
    9·1 answer
  • I'll. ark brainlyest and give a lot of points for answering this right ​
    9·1 answer
  • Solve for x. <br><br> HINT: these are exterior angles
    11·2 answers
  • What is the value of the sum 4 + 9
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!