~You would set up the equation x+(3x+5)=6565
~Combine like products so 3x+x=4x.
~Now you have 4x+5=6565.
~Isolate the variable by subracting the 5 from both sides, so you have 4x=6560.
~Since you multiply x by 4, divide both sides by 4, and 6560 divided by 4 is 1,640.
*The first number is 1,640.
~Now to find the second number you simply plug in 1,640 to (3x+5), with 1,640 being x.
~That would become [3(1,640)+5].
~3 times 1, 640 is 4,920, and add the 5 so you get 4,925.
*The second and larger number is 4,925.
If you wanna check just add the two numbers, and you get 6565.
Answer:
16.
Step-by-step explanation:
The whole numbers stay the same so it stays as 2. The way they get 30 for the denominator is by multiplying the fraction 1/3 by 10 and 1/5 by 6. So now we have 10/30 and 6/30. Add them together, 10+6=16
Answer:
Step-by-step explanation:
what grade are you on lemme help you
So you want to solve for x?
It would be nice if this would easily factor:
(-4x + 5)(2x +1) = 0 This will not work!
So you need to use the quadratic formula:
a = -8, b = 4, c = 5

x = (-4 +/-

)/2(-8)
= (-4 +/-

)/-16
= (-4 +

)/-16
= 1/4 -

/4
<u><em>Answer:</em></u>
Part a .............> x = 11
Part b .............> k = 57.2
Part c .............> y = 9.2
<u><em>Explanation:</em></u>
The three problems deal with inverse variation between two variables
An inverse variation relation between two variables means that when one of the variables increases, the other will decrease (and vice versa)
<u>Mathematically, an inverse variation relation is represented as follows:</u>

where x and y are the two variables and k is the constant of variation
<u><em>Now, let's check the givens:</em></u>
<u>Part a:</u>
We are given that y = 3 and k = 33
<u>Substitute in the original relation and solve for x as follows:</u>

<u>Part b:</u>
We are given that y = 11 and x = 5.2
<u>Substitute in the original relation and solve for k as follows:</u>

<u>Part c:</u>
We are given that x=7.8 and k=72
<u>Substitute in the original relation and solve for y as follows:</u>
to the nearest tenth
Hope this helps :)