Thomas has $6.35 in dimes and quarters. The number of dimes is three more than three times the number of quarters. If q represents the number of quarters, then which of the following expressions represents the number of dimes that Thomas has? 3d + 3 3q + 3 6.35 - q 10(3q + 3)

Answers

Answer 1
Answer: 3q+3 because if the dimes is 3x as many as the quarter and plus 3 then you would multiply the quarters by 3 then add 3 and that would give you the number of dimes


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Andrew’s rotation maps point M(9, –1) to M’(–9, 1). Which describes the rotation?

Answers

180 rotation  i took the test all the other ones were wrong                                                                                                                

Answer:

180

Step-by-step explanation:

Solve for k:
2k(7-5k)+11=6k+3(k^2-1)
with work please.

Answers

2k(7-5k)+11=6k+3(k^2-1) 
⇒ (2k)* 7 - (2k)* (5k) +11= 6k+ 3* k^2- 3*1 (distributive property)
⇒ 14k -10k^2 +11= 6k+ 3k^2 -3
⇒ (14k -14k) +(-10k^2 +10k^2)+ (11-11) = (6k-14k)+ (3k^2+10k^2)+ (-3-11)
⇒ 13k^2- 8k- 14= 0
⇒ 13 (k^2- 8/13k- 14/13)=0
⇒ (k- ((4+3*sqrt(22))/13))* ((k- ((4-3*sqrt(22))/13))= 0
k= (4+3*sqrt(22))/13 or k= (4-3*sqrt(22))/13).

Hope this helps~

Answer:

k= (4+3*sqrt(22))/13 or k= (4-3*sqrt(22))/13).

Step-by-step explanation:

2k(7-5k)+11=6k+3(k^2-1)

We move all terms to the left:

2k(7-5k)+11-(6k+3(k^2-1))=0

We add all the numbers together, and all the variables

2k(-5k+7)-(6k+3(k^2-1))+11=0

We multiply parentheses

-10k^2+14k-(6k+3(k^2-1))+11=0

We calculate terms in parentheses: -(6k+3(k^2-1)), so:

6k+3(k^2-1)

We multiply parentheses

3k^2+6k-3

Back to the equation:

-(3k^2+6k-3)

We get rid of parentheses

-10k^2-3k^2+14k-6k+3+11=0

We add all the numbers together, and all the variables

-13k^2+8k+14=0

a = -13; b = 8; c = +14;

Δ = b2-4ac

Δ = 82-4·(-13)·14

Δ = 792

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

k1=−b−Δ√2ak2=−b+Δ√2a

The end solution:

Δ−−√=792−−−√=36∗22−−−−−−√=36−−√∗22−−√=622−−√

k1=−b−Δ√2a=−(8)−622√2∗−13=−8−622√−26

k2=−b+Δ√2a=−(8)+622√2∗−13=−8+622√−26

   

   

The Main Street Cinema holds 1,356 people. Last Saturday night, they were sold out for two showings of Super Alien 5. There was a special price of $5 for every seat for each showing. How much money did the theater take in that night? A. $10,888 B. $9,848 C. $6,780 D. $13,560

Answers

The amount the theater take in that night is (d) $13560

How to determine the amount the theater take in that night?

From the question, we have the following parameters that can be used in our computation:

People = 1356

Number of shows = 2

Price per ticket = $5

using the above as a guide, we have the following:

Total = People * Number of shows * Price per ticket

So, we have

Total = 1356 * 2 * 5

Evaluate

Total = 13560

Hence, the amount the theater take in that night is (d) $13560

Read more about expressions at

brainly.com/question/34777630

#SPJ6

1356 x 2 = n 
n x 5 = Answer

2712 x 5 = $13,560

D

A Pythagorean triple is a triple of natural numbers satisfying the equation a^2+b^2+c^2.One way to produce a Pythagorean triple is to add the reciprocals of any two consecutive even or odd numbers. For example, 1/5+1/7=12/35. Now 12^2+35^2=1369. This is a Pythagorean triple if 1369 is a perfect square, which it is since 1369=37^2. So 12, 35, 37 is a Pythagorean triple. Prove that this method always works.

Answers

x, x+2 - two consecutive odd or even numbers
Add the reciprocals of these numbers.
(1)/(x)+(1)/(x+2)=(x+2)/(x(x+2))+(x)/(x(x+2))=(x+2+x)/(x^2+2x)=(2x+2)/(x^2+2x)

Now add the squares of the numerator and denominator, as in the example.
(2x+2)^2+(x^2+2x)^2= \n 4x^2+8x+4+x^4+4x^3+4x^2= \n x^4+4x^3+8x^2+8x+4

So this number has to be a perfect square.
x^4+4x^3+8x^2+8x+4= \nx^4+2x^3+2x^2+2x^3+4x^2+4x+2x^2+4x+4= \nx^2(x^2+2x+2)+2x(x^2+2x+2)+2(x^2+2x+2)= \n(x^2+2x+2)(x^2+2x+2)= \n(x^2+2x+2)^2
It is a perfect square, so this method always works.

The numbers 2x+2, \ x^2+2x, \ (x^2+2x+2)^2 are a Pythagorean triple for any x \in \mathbb{N^+}.

Answer:

even tho this has nothing to do with the answer ;-;

Step-by-step explanation:First a definition: A Pythagorean Triple are three natural numbers 1 <= a <= b <= c, such that a2 + b2 = c2 holds. For example 3, 4, 5 is such a triple, since 32 + 42 = 9 + 16 = 25 = 52. While 2, 3, 4 is not such a triple, since 22 + 32 = 4 + 9 = 13 and 42 = 16. We note here that only natural numbers are considered, and thus 2, 3 can not be extended to Pythagorean triple (since 13 is not the square of some integer).

Now the question: Can we colour the natural numbers 1, 2, 3, ... with two colours, say blue and red, such that there is no monochromatic Pythagorean triple? In other words, is it possible to give every natural number one of the colours blue or red, such that for every Pythagorean triple a, b, c at least one of a, b, c is blue, and at least one of a, b, c is red ? We prove: The answer is No. That is easier to express positively: Whenever we colour the natural numbers blue or red, there must exist a monochromatic triple (one blue triple or one red triple).

More precisely we prove, using "bi-colouring" for colouring blue or red: 1) However we bi-colour the numbers 1, ..., 7825, there must exist a monochromatic Pythagorean triple. 2) While there exists a bi-colouring of 1, ..., 7824, such that no Pythagorean triple is monochromatic. Part 2) is relatively easy. Part 1) is the real hard thing -- every number from 1, ..., 7825 gets one of two possible colours, so altogether there are 27825 possible colourings, which all in a sense need to be considered, and need to be excluded. What is 27825? It is approximately 3.63 * 102355, that is, a number with 2356 decimal places. The number of particles in the universe is at most 10100, a tiny number with just 100 decimal places (in comparison).

Now let's perform real brute-force, running through all the possibilities, one after another: Even if we could place on every particle in the universe a super-computer, and they all would work perfectly together for the whole lifetime of the universe -- by far not enough. Even not if inside every particle we could place a whole universe. Even if each particle in the inner universe becomes again itself a universe, with every particle carrying a super-computer, still

by far not enough. Hope you get the idea -- the $100 we got wouldn't pay that energy bill.

Fortunately there comes SAT solving to the rescue, which actually is really good with such tasks -- it can solve some such task and even more monstrous tasks. Our ``brute-reasoning'' approach solved the problem and resulted into a 200 terabytes proof -- the largest math proof ever. Though we must emphasise that this is in no way guaranteed, and possibly it will take aeons! SAT solving uses propositional logic, in the special form of CNF (conjunctive normal form). Fortunately, in this case it is easy to represent our problem in this form.

Six more than fives times a number (x) is at least 21. I need an inequality.

Answers

5n + 6 ≥ 21 would be the way to write this in math language.

It says that 5n + 6 is 21 or is more than 21.  The symbol ≥ says 21 or more in this case.

. What is the slope of the line with the equation 2x + 3y + 6 = 0 ?

Answers

-2/3. You take 3y and move it to the opposite side. Then you would take the three from the 3Y and divide everything by three, so you get -2/3x+2=Y then you look at what the coefficient is in front of X and X equals -2/3, which is your slope.
2x + 3y + 6 = 0 \n \n3y=-2x-6 /:3\n \ny=-(2)/(3)x-(6)/(3)\n \ny=-(2)/(3)x-2

Answer : \ is \ your \ slope : \ \ m = -(2)/(3)