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avanturin [10]
3 years ago
11

2x + 5y = 17 what is x what is y

Mathematics
1 answer:
andrew-mc [135]3 years ago
8 0

Answer:

x = (17 - 5y)/2

y = (17 - 2x)/5

Step-by-step explanation:

nothing else can be done here.

if you are looking for actual numbers as solution, then you have forgotten to put something important here.

to solve a system with 2 variables we need 2 equations. it cannot be solved with only 1 equation.

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The following shapes are similar. Set up equation and solve for the variables
koban [17]

Answer:

x=10

Step-by-step explanation:

For number 19.

The equation you get is:

X+2=6(2) this is because 5x2=10, so you have to do the same here.

so, X+2=12

-2 from both sides and you get

x=10

5 0
3 years ago
One number is 3 times another. If the sum of the two numbers is 15, find the two numbers.
gregori [183]

Answer:

15/4 and 45/4

Step-by-step explanation:

Let x and y be the numbers

x = 3y

x+ y = 15

Substitute the first equation in to the second equation

3y+y = 15

4y = 15

y = 15/4

x = 3(15/4)

x = 45/4

The two numbers are 15/4 and 45/4

8 0
3 years ago
Read 2 more answers
Arrange the steps from start to finish to factor this expression using the grouping method x^2-5x-24 drag the tiles to the corre
rosijanka [135]

Answer:

(x - 8)(x + 3)

Step-by-step explanation:

x² - 5x - 24

Consider the factors of - 24 which sum to give the coefficient of the x- term (- 5)

The factors are - 8 and + 3 , since

- 8 × + 3 = - 24 and - 8 + 3 = - 5

Use these factors to split the x- term

x² - 8x + 3x - 24 ( factor the first/second and third/fourth terms )

= x(x - 8) + 3(x - 8) ← factor out (x - 8) from each term

= (x - 8)(x + 3) ← in factored form

7 0
2 years ago
Solve by factoring:<br><br> 5x^2 + 19x + 12 = 0
Mariulka [41]

When solving polynomial equations, first set the equation

equal to zero, then check for a Greatest Common Factor.

Since there is no greatest common factor, this will make life a

little bit more difficult because we will have to factor the trinomial.

First, find all possible factors for constant term.

I have listed these in red below.

Make sure to reverse the factors since in each binomial,

the first term is different so we need to reverse the factors.

The factors of 5x² will just be 5x and z.

After finding factors for the constant and the leading

coefficient for your trinomial, we can eliminate some factors.

If you have an odd coefficient on the middle term,

you can eliminate any pairs of even factors.

So the group I crossed out in purple below will be eliminated.

Another strategy is to use factors a little closer together.

So I would go with the 4 and 3 group.

Surely enough, our trinomial does factor with the 4 and 3.

Now set the factored trinomial equal to 0.

Then use zero product property and you have your answer.

5 0
3 years ago
Read 2 more answers
A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t
MakcuM [25]

Answer:

165.

Step-by-step explanation:

Since repetition isn't allowed, there would be 11 choices for the first donut, (11 - 1) = 10 choices for the second donut, and (11 - 2) = 9 choices for the third donut. If the order in which donuts are placed in the bag matters, there would be 11 \times 10 \times 9 unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type 3 \times 2 \times 1 = 6 times.

For example, if a bag includes donut of type x, y, and z, the count 11 \times 10 \times 9 would include the following 3 \times 2 \times 1 arrangements:

  • xyz.
  • xzy.
  • yxz.
  • yzx.
  • zxy.
  • zyx.

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count 11 \times 10 \times 9 by 3 \times 2 \times 1 = 6 to find the actual number of donut combinations:

\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}.

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of 3 objects from a set of 11 distinct objects:

\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}.

5 0
2 years ago
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