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3 years ago

The sum of two consecutive odd numbers is 284. Find each number please.

2 answers:

8 0

We can solve this by finding one of the numbers first
Let the smaller number be y.
Since these 2 numbers are consecutive odd numbers, the larger number should be 2 more than the smaller one (y+2).

Therefore,
y + (y+2) = 284
2y +2 = 284
shift +2 to the other side and turn it into -2
2y = 284 - 2
2y = 282
shift x2 to the other side and turn it into /2
y = 282/2
y = 141

Now we got the smaller number, which is 141, we can also find the larger number by adding 2 to it. 141+2 = 143,
Therefore, your answer is 141 and 143

andrey2020 [161]3 years ago

3 0

We need to understand how to set up the equation to solve this. If they are consecutive odd integers, that means they're two odd numbers that are next to each other, like 3 and 5 or 27 and 29. There's always a two-integer difference between the two numbers since they're both odd, so we'll use that to set up our equation. Let's make the smaller integer x. That will make the larger integer x+2, since there is a difference between them of 2. That means our equation is x+x+2=284 or 2x+2=284. When we subtract 2 from each side and then divide by two, we'll get x=141, the first integer. That means the two integers are 141 and 143.

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Can someone check this for me please?

anyanavicka [17]

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3 years ago

PLEASE HELP ME! I am suck on this math assignment

patriot [66]

Hey there! I'm happy to help!

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QUESTION 18

Here is how you multiply a 2×2 matrix by a column vector (just one column matrix).

$\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] *\left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}ax&by\\cx&dy\\\end{array}\right]$

Once you get that final matrix, you add the rows which will give you a final column vector. Let's do it with this question, as we are multiplying together the matrices A and v.

$\left[\begin{array}{ccc}-1&4\\2&0\\\end{array}\right] *\left[\begin{array}{ccc}8\\6\\\end{array}\right] =\left[\begin{array}{ccc}-8&24\\16&0\\\end{array}\right] =\left[\begin{array}{ccc}16\\16\\\end{array}\right]$

So, your answer is $\left[\begin{array}{ccc}16\\16\\\end{array}\right]$ .

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QUESTION 14

We have this equation of an ellipse and we need to solve for y. When we have a more complex equation to solve you get rid of the things furthest from the variable you are solving for first and then you get closer and closer to it.

$\frac{x^2}{25} + \frac{y^2}{16}=1$

We subtract $\frac{x^2}{25}$ from both sides.

$\frac{y^2}{16}=1-\frac{x^2}{25}$

We multiply both sides by 16.

$y^2=1-\frac{x^2}{25}*16$

We square root both sides.

$y=$ $\sqrt{1-\frac{x^2}{25}*16$

We can take a four out of our root (because 16 is a perfect square so it can be simplified more).

y= $4\sqrt{1-\frac{x^2}{25}$

And we take out 1/5 (which is the square root of 1/25).

y=± $\frac{4}{5}\sqrt{1-x^2$

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QUESTION 17

We see that Sea Slugs have 3/10 matches won, so they have won 30% of their matches.

From now on, they will be winning 2/3 of their games. Let's look at the answer options.

A. If they play 15 more games, they will win 10 of them. Our total matches will be 25 (our original 10+15). So, we will have 13/25, which is equal to 52%, so this is incorrect.

B. If they play 10 more games they win 6 2/3 games (like that makes any sense) for a total of 9 2/3 games out of 20. This is about 48%, so this is incorrect.

C. If they play 40 games, they are winning fractional games again. They win 29 2/3 games out of 50 total, which is about 59%, but not quite 60. This is also incorrect.

D. If they play 45 matches, they will win 30 games. Add the original 10 and you have 40 games. Our total is 55. 40/55 is about 73%, so they have passed that 60% threshold, so this is the correct answer.

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QUESTION 16

Our question tells us that if we have two numbers x and y, and we add them, it has to be 24 or greater. By saying it can be equal to 24 or greater, we are going to be using a ≤ or ≥ sign. The opening of the arrow always faces the equal or greater quantity, and in this case it is x+y, so it will look like x+y≥24.

This eliminates answer choices B and D as they do not have the equal part on the inequality.

Product means to multiply, so it will be written as xy. It has to be less than 60, so in this case the greater quantity is 60, so the arrow will opening towards 60, so it will look like xy<60.

So, our final answer is A.

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QUESTION 12

You can only multiply matrices if the number of columns in the first matrix is equal to the number of rows in the second. So basically the two interior numbers have to be the same (a,b) and (b,c). So, if we follow this, we get these answers:

2×2 times 2×1 YES

2×1 times 2×2= NO

3×3 times 3×1= YES

3×1 times 3×3= NO

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Have a wonderful day and keep on learning! :D

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