katie has 60 flowers. she wants to divide into two groups so the ratio is 2:3 how many flowers in each group

I need help, please!

Neporo4naja [7]

Answer:

2 of these numbers are divisible by 4 and 5.

20 and 60 are divisible by 4 and 5.

Step-by-step explanation:

20 is divisible by 4 and 5.

Divisible just means that the 4 (also the 5) goes into 20 evenly, with no remainder.

4 goes into 20 evenly with no remainder.

5 goes into 20 evenly with no remainder.

4 does not go into 35 evenly with no remainder.

35 ÷ 4 is 8 with a remainder of 3.

35 ÷ 4 = 8.75

5 does go into 35. 35 is divisible by 5.

35 is not divisible by 4.

60 is divisible by 4

60 ÷ 4 = 15 (no remainder, no extra decimals if you do it on a calculator)

60 ÷ 5 = 12 (no remainder, no extra decimals if you do it on a calculator)

60 is divisible by 4 and also by 5.

3 0

3 years ago

The system of equations below has no solution.

mixer [17]

Answer:

Second option: 0 = 26

Explanation:

This is the given system of equations:

$\frac{2}{3} x+\frac{5}{2} y=15\\ \\ 4x+15y=12$

A linear combination of the system is any equation formed by the algebraic addition of both equations, one or both multiplied by an arbitrary constant.

To prove that the given system has no solution you could multiply the first equation times 6 (to get rid of the fractions), multiply the second equation times - 1, and add the two results:

1. First equation times 6:

$6\times\frac{2}{3} x+6\times\frac{5}{2}y=6\times 15\\ \\ 4x+15y=90$

2. Second equation times - 1:

$-4x-15y=-12$

3. Add the two new equations:

$0=78$

4. Conclusion:

Since 0 = 78 is false, no matter what the value of x and y are, the conclusion is that the system of equations has not solution.

The only choice that represents that same situation is the second one, 0 = 26. That is a possible linear combination that represents that the system of equations has no solutions.

In fact, you might calculate the exact factors by which you had to multiply each one of the original equations to get 0 = 26, but it is not necessary to tell that that option represents a possible linear combination for the given system of equations.

7 0

3 years ago

Read 2 more answers

10 red cards and 10 and black cards are placed in a bag. You choose one card and then another without replacing the first card.

MrMuchimi

2/10 right ? hope this helped :)

4 0

3 years ago

Divide. give the quotient and remainder 295/4

Sonja [21]

Answer:73 and remainder of 3

Step-by-step explanation:

295/4=73 and a remiander of 3

6 0

2 years ago

Read 2 more answers

Solve the following equations for x,

stellarik [79]

(i) 3 csc²(x) - 4 = 0

3 csc²(x) = 4

csc²(x) = 4/3

sin²(x) = 3/4

sin(x) = ± √3/2

x = arcsin(√3/2) + 2nπ or x = arcsin(-√3/2) + 2nπ

x = π/3 + 2nπ or x = -π/3 + 2nπ

where n is any integer. The general result follows from the fact that sin(x) is 2π-periodic.

In the interval 0 ≤ x ≤ 2π, the first family of solutions gives x = π/3 and x = 4π/3 for n = 0 and n = 1, respectively; the second family gives x = 2π/3 and x = 5π/3 for n = 1 and n = 2.

(ii) 4 cos²(x) + 2 cos(x) - 2 = 0

2 cos²(x) + cos(x) - 1 = 0

(2 cos(x) - 1) (cos(x) + 1) = 0

2 cos(x) - 1 = 0 or cos(x) + 1 = 0

2 cos(x) = 1 or cos(x) = -1

cos(x) = 1/2 or cos(x) = -1

[x = arccos(1/2) + 2nπ or x = 2π - arccos(1/2) + 2nπ] or x = arccos(-1) + 2nπ

[x = π/3 + 2nπ or x = 5π/3 + 2nπ] or x = π + 2nπ

For 0 ≤ x ≤ 2π, the solutions are x = π/3, x = 5π/3, and x = π.

7 0

3 years ago